Deck 17: Second-Order Differential Equations

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سؤال
A spring with a mass of 2 kg has damping constant 14, and a force of A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. The spring is stretched 1m beyond its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t.<div style=padding-top: 35px> N is required to keep the spring stretched A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. The spring is stretched 1m beyond its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t.<div style=padding-top: 35px> m beyond its natural length. The spring is stretched 1m beyond its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t.
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سؤال
Solve the differential equation using the method of variation of parameters. y+4y+4y=e2xx3y ^ { \prime \prime } + 4 y ^ { \prime } + 4 y = \frac { e ^ { - 2 x } } { x ^ { 3 } }

A) y(x)=e2x(c1+c212x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } \frac { 1 } { 2 x } \right)
B) y(x)=e2x(c1+c2x+12x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x + \frac { 1 } { 2 x } \right)
C) y(x)=e2x(c1+c2)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } \right)
D) y(x)=e2x(c1+c2x)+12xy ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x \right) + \frac { 1 } { 2 x }
E) y(x)=e2x(c1+c2x+c312x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x + c _ { 3 } \frac { 1 } { 2 x } \right)
سؤال
Use power series to solve the differential equation.. (x2+1)y+xyy=0\left( x ^ { 2 } + 1 \right) y ^ { \prime \prime } + x y ^ { \prime } - y = 0

A) y(x)=c0n=2(1)n1(2n3)!22n2n!(n2)!x2ny ( x ) = c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n }
B) y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2n!(n2)!x2n+1y ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n + 1 }
C) y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2(n2)!x2ny ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } ( n - 2 ) ! } x ^ { 2 n }
D) y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2n!(n2)!x2ny ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n }
E) y(x)=c0+c1x+c0x22+c0n=2(1)n+1(2n)!22n+2n!(n2)!x2n+1y ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } ( 2 n ) ! } { 2 ^ { 2 n + 2 } n ! ( n - 2 ) ! } x ^ { 2 n + 1 }
سؤال
Use power series to solve the differential equation. y=25yy ^ { \prime \prime } = 25 y

A) y(x)=c1cosh5x+c2sinh5xy ( x ) = c _ { 1 } \cosh 5 x + c _ { 2 } \sinh 5 x
B) y(x)=c15coshx+c25sinhx+exy ( x ) = c _ { 1 } 5 \cosh x + c _ { 2 } 5 \sinh x + e ^ { x }
C) y(x)=5coshx+5sinhxy ( x ) = 5 \cosh x + 5 \sinh x
D) y(x)=c1cosh5x+c2sinh5x+exy ( x ) = c _ { 1 } \cosh 5 x + c _ { 2 } \sinh 5 x + e ^ { x }
E) y(x)=c1cosh5x+c2sinh5x+x2y ( x ) = c _ { 1 } \cosh 5 x + c _ { 2 } \sinh 5 x + x ^ { 2 }
سؤال
A series circuit consists of a resistor R=32ΩR = 32 \Omega , an inductor with L=4HL = 4 H , a capacitor with C=0.003125 FC = 0.003125 \mathrm {~F} , and a 2424 -V battery. If the initial charge is 0.0008 C and the initial current is 0, find the current I(t) at time t.

A) I(t)=0.742e4tsin(8t)I ( t ) = 0.742 e ^ { - 4 t } \sin ( 8 t )
B) I(t)=e4t160cos(8t)I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t )
C) I(t)=e4t160cos(8t)12I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t ) - \frac { 1 } { 2 }
D) I(t)=0.742e8tI ( t ) = 0.742 e ^ { - 8 t }
E) I(t)=e4t160(0.003125cos(4t)+8sin(4t))12I ( t ) = \frac { e ^ { - 4 t } } { 160 } ( 0.003125 \cos ( 4 t ) + 8 \sin ( 4 t ) ) - \frac { 1 } { 2 }
سؤال
Use power series to solve the differential equation. y=7x3yy ^ { \prime } = 7 x ^ { 3 } y

A) y(x)=7ex44y ( x ) = 7 e ^ { \frac { x ^ { 4 } } { 4 } }
B) y(x)=ce7xy ( x ) = c e ^ { 7 x }
C) y(x)=ex44y ( x ) = e ^ { - \frac { x ^ { 4 } } { 4 } }
D) y(x)=ce7x44y ( x ) = c e ^ { - \frac { 7 x ^ { 4 } } { 4 } }
E) y(x)=ce7x44y ( x ) = c e ^ { \frac { 7 x ^ { 4 } } { 4 } }
سؤال
A series circuit consists of a resistor A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t.<div style=padding-top: 35px> an inductor with L = A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t.<div style=padding-top: 35px> H, a capacitor with
C = A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t.<div style=padding-top: 35px> F, and a A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t.<div style=padding-top: 35px> -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t.
سؤال
A spring has a mass of 11 kg and its damping constant is c=10c = 10 . The spring starts from its equilibrium position with a velocity of 11 m/s. Graph the position function for the spring constant k=20k = 20 .

A)  <strong>A spring has a mass of  1  kg and its damping constant is  c = 10  . The spring starts from its equilibrium position with a velocity of  1  m/s. Graph the position function for the spring constant  k = 20  .</strong> A)   B)   C)   <div style=padding-top: 35px>
B)  <strong>A spring has a mass of  1  kg and its damping constant is  c = 10  . The spring starts from its equilibrium position with a velocity of  1  m/s. Graph the position function for the spring constant  k = 20  .</strong> A)   B)   C)   <div style=padding-top: 35px>
C)  <strong>A spring has a mass of  1  kg and its damping constant is  c = 10  . The spring starts from its equilibrium position with a velocity of  1  m/s. Graph the position function for the spring constant  k = 20  .</strong> A)   B)   C)   <div style=padding-top: 35px>
سؤال
Use power series to solve the differential equation. (x10)y+2y=0( x - 10 ) y ^ { \prime } + 2 y = 0

A) y(x)=c0n=0xn10ny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 10 ^ { n } }
B) y(x)=c0n=0n+110nxny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { n + 1 } { 10 ^ { n } } x ^ { n }
C) y(x)=c0n=0(n+1)xny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { ( n + 1 ) } { x ^ { n } }
D) y(x)=c0n=0(n+1)xny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } ( n + 1 ) x ^ { n }
E) y(x)=c0n=0nxn10ny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { n x ^ { n } } { 10 ^ { n } }
سؤال
A spring with a 1616 -kg mass has natural length 0.80.8 m and is maintained stretched to a length of 1.21.2 m by a force of 19.619.6 N. If the spring is compressed to a length of 0.40.4 m and then released with zero velocity, find the position x(t)x ( t ) of the mass at any time tt .

A) x(t)=0.4e4tcos(0.4t)x ( t ) = - 0.4 e ^ { - 4 t } \cos ( 0.4 t )
B) x(t)=0.8sin(1.75t)x ( t ) = 0.8 \sin ( 1.75 t )
C) x(t)=0.4e4tcos(1.75t)x ( t ) = - 0.4 e ^ { - 4 t } \cos ( 1.75 t )
D) x(t)=0.4cos(1.75t)+0.4sin(1.75t)x ( t ) = - 0.4 \cos ( 1.75 t ) + 0.4 \sin ( 1.75 t )
E) x(t)=0.4cos(1.75t)x ( t ) = - 0.4 \cos ( 1.75 t )
سؤال
Suppose a spring has mass M and spring constant k and let ω=k/M\omega = \sqrt { k / M } . Suppose that the damping constant is so small that the damping force is negligible. If an external force F(t)=8F0cos(ωt)F ( t ) = 8 F _ { 0 } \cos ( \omega t ) is applied (the applied frequency equals the natural frequency), use the method of undetermined coefficients to find the equation that describes the motion of the mass.

A) x(t)=c1cos(ωt)+c2sin(ωt)+F0t22Mωsin(ωt)x ( t ) = c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) + \frac { F _ { 0 } t ^ { - 2 } } { 2 M \omega } \sin ( \omega t )
B) x(t)=c1cos(ωt)+c2sin(ωt)+F0eωt2Mωx ( t ) = c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) + \frac { F _ { 0 } e ^ { - \omega t } } { 2 M \omega }
C) x(t)=c1cos(ωt)+c2sin(ωt)+4F0tMωsin(ωt)x ( t ) = c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) + \frac { 4 F _ { 0 } t } { M \omega } \sin ( \omega t )
D) x(t)=F0t22Mωcos(ωt)x ( t ) = \frac { F _ { 0 } t ^ { 2 } } { 2 M \omega } \cos ( \omega t )
E) x(t)=F0t2Mω(c1cos(ωt)+c2sin(ωt))x ( t ) = \frac { F _ { 0 } t } { 2 M \omega } \left( c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) \right)
سؤال
A spring with a mass of 2 kg has damping constant 14, and a force of A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. Find the mass that would produce critical damping.<div style=padding-top: 35px> N is required to keep the spring stretched A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. Find the mass that would produce critical damping.<div style=padding-top: 35px> m beyond its natural length. Find the mass that would produce critical damping.
سؤال
A spring with a mass of 99 kg has damping constant 28 and spring constant 195195 . Find the damping constant that would produce critical damping.

A) c=c = 3619536 \sqrt { 195 }
B) c=c = 23402340
C) c=1959c = 195 \sqrt { 9 }
D) c=c = 9 195\sqrt { 195 }
E) c=c = 61956 \sqrt { 195 }
سؤال
Use power series to solve the differential equation. yxyy=0,y(0)=7,y(0)=0y ^ { \prime \prime } - x y ^ { \prime } - y = 0 , y ( 0 ) = 7 , y ^ { \prime } ( 0 ) = 0

A) y(x)=7exy ( x ) = 7 e ^ { x }
B) y(x)=7ex22y ( x ) = 7 e ^ { - \frac { x ^ { 2 } } { 2 } }
C) y(x)=e7xy ( x ) = e ^ { 7 x }
D) y(x)=cex22y ( x ) = c e ^ { \frac { x ^ { 2 } } { 2 } }
E) y(x)=7ex22y ( x ) = 7 e ^ { \frac { x ^ { 2 } } { 2 } }
سؤال
A spring with a 3-kg mass is held stretched 0.9 m beyond its natural length by a force of 30 N. If the spring begins at its equilibrium position but a push gives it an initial velocity of 11 m/s, find the position x(t) of the mass after t seconds.

A) x(t)=1sin(7t)x ( t ) = 1 \sin ( 7 t )
B) x(t)=0.3sin(103t)+0.4cos(103t)x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right) + 0.4 \cos \left( \frac { 10 } { 3 } t \right)
C) x(t)=0.3sin(103t)x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right)
D) x(t)=0.4cos(103t)x ( t ) = 0.4 \cos \left( \frac { 10 } { 3 } t \right)
E) x(t)=103sin(7t)x ( t ) = \frac { 10 } { 3 } \sin ( 7 t )
سؤال
The solution of the initial-value problem x2y+xy+x2y=0,y(0)=1,y(0)=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is called a Bessel function of order 0. Solve the initial - value problem to find a power series expansion for the Bessel function.

A) y(x)=n=0(1)n+1nx2n122n1(n!)2y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { n x ^ { 2 n - 1 } } { 2 ^ { 2 n - 1 } ( n ! ) ^ { 2 } }
B) y(x)=n=0(1)nxn2n(2n!)y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { n } } { 2 ^ { n } ( 2 n ! ) }
C) y(x)=n=0(1)nx2n22n(n!)2y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { 2 ^ { 2 n } ( n ! ) ^ { 2 } }
D) y(x)=2x+n=0(1)nx2n(n!)2y ( x ) = 2 ^ { x } + \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { ( n ! ) ^ { 2 } }
E) y(x)=x+n=0(1)nx2n22n(n!)2y ( x ) = x + \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { 2 ^ { 2 n } ( n ! ) ^ { 2 } }
سؤال
A series circuit consists of a resistor R=96ΩR = 96 \Omega , an inductor with L=8HL = 8 \mathrm { H } , a capacitor with C=0.00125 FC = 0.00125 \mathrm {~F} , and a generator producing a voltage of E(t)=48cos(10t)E ( t ) = 48 \cos ( 10 t ) If the initial charge is Q=0.001CQ = 0.001 \mathrm { C } and the initial current is 0, find the charge Q(t)Q ( t ) at time t.

A) Q(t)=e6t(0.001cos(8t)0.06175sin(8t))+120sin(10t)Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 8 t ) - 0.06175 \sin ( 8 t ) ) + \frac { 1 } { 20 } \sin ( 10 t )
B) Q(t)=e2t(0.051cos(2t)0.06175sin(2t))+120sin(10t)Q ( t ) = e ^ { - 2 t } ( 0.051 \cos ( 2 t ) - 0.06175 \sin ( 2 t ) ) + \frac { 1 } { 20 } \sin ( 10 t )
C) Q(t)=e6t(cos(6t)0.06175sin(6t))Q ( t ) = e ^ { - 6 t } ( \cos ( 6 t ) - 0.06175 \sin ( 6 t ) )
D) Q(t)=e3t(cos(5t)0.06175sin(5t))Q ( t ) = e ^ { - 3 t } ( \cos ( 5 t ) - 0.06175 \sin ( 5 t ) )
E) Q(t)=e6t(0.001cos(8t)0.06175sin(8t))Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 8 t ) - 0.06175 \sin ( 8 t ) )
سؤال
A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.

A) <strong>A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.</strong> A)   B)   C)   <div style=padding-top: 35px>
B) <strong>A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.</strong> A)   B)   C)   <div style=padding-top: 35px>
C) <strong>A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.</strong> A)   B)   C)   <div style=padding-top: 35px>
سؤال
The figure shows a pendulum with length L and the angle θ\theta from the vertical to the pendulum. It can be shown that θ\theta , as a function of time, satisfies the nonlinear differential equation d2θdt2+gLsinθ=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0 where g=9.8 m/s2g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }  is the acceleration due to gravity. For small values of \text { is the acceleration due to gravity. For small values of } θ\theta we can use the linear approximation sinθ=θ\sin \theta = \theta  and then the differential equation becomes linear. Find the equation \text { and then the differential equation becomes linear. Find the equation }  of motion of a pendulum with length 1 m if θ is initially 0.2rad and the initial \text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }  angular velocity is \text { angular velocity is } dθdt=1rad/s\frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }  <strong>The figure shows a pendulum with length L and the angle  \theta  from the vertical to the pendulum. It can be shown that  \theta  , as a function of time, satisfies the nonlinear differential equation  \frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0  where  g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }   \text { is the acceleration due to gravity. For small values of }   \theta  we can use the linear approximation  \sin \theta = \theta   \text { and then the differential equation becomes linear. Find the equation }   \text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }   \text { angular velocity is }   \frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }   </strong> A)  \theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )  B)  \theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )  C)  \theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )  D)  \theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )  E)  \theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )  <div style=padding-top: 35px>

A) θ(t)=0.2cos(9.8t)+19.8sin(9.8t)\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )
B) θ(t)=0.2cos(9.8t)+2sin(9.8t)\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )
C) θ(t)=2cos(9.8t)+19.8sin(9.8t)\theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )
D) θ(t)=19.8cos(9.8t)+0.2sin(9.8t)\theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )
E) θ(t)=0.2sin(9.8t)+19.8cos(9.8t)\theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )
سؤال
Use power series to solve the differential equation. y+x2y=0,y(0)=6,y(0)=0y ^ { \prime \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { \prime } ( 0 ) = 0

A) y(x)=6n=1(1)nx4n4n(4n1)(4n4)(4n5)43y ( x ) = 6 - \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
B) y(x)=n=1(1)n6x4n4n(4n1)(4n4)(4n5)43y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
C) y(x)=6+n=1(1)n6x4n4n(4n1)(4n4)(4n5)43y ( x ) = 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
D) y(x)=6+n=1(1)nx4n4n(4n1)(4n4)(4n5)43y ( x ) = - 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
E) y(x)=n=1(1)nx4n4n(4n1)(4n4)(4n5)43y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
سؤال
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of undetermined coefficients. Solve the differential equation using the method of undetermined coefficients.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of undetermined coefficients. y3y=sin9xy ^ { \prime \prime } - 3 y ^ { \prime } = \sin 9 x

A) y(x)=c1e3x+1270cos(6x)190sin(6x)y ( x ) = c _ { 1 } e ^ { 3 x } + \frac { 1 } { 270 } \cos ( 6 x ) - \frac { 1 } { 90 } \sin ( 6 x )
B) y(x)=c1+c2e3x+1270cos(6x)190sin(6x)y ( x ) = c _ { 1 } + c _ { 2 } e ^ { 3 x } + \frac { 1 } { 270 } \cos ( 6 x ) - \frac { 1 } { 90 } \sin ( 6 x )
C) y(x)=c1e3x+190cos(6x)+1270sin(6x)y ( x ) = c _ { 1 } e ^ { 3 x } + \frac { 1 } { 90 } \cos ( 6 x ) + \frac { 1 } { 270 } \sin ( 6 x )
D) y(x)=c1+c2e3x190sin(6x)y ( x ) = c _ { 1 } + c _ { 2 } e ^ { 3 x } - \frac { 1 } { 90 } \sin ( 6 x )
E) y(x)=1270cos(6x)190sin(6x)y ( x ) = \frac { 1 } { 270 } \cos ( 6 x ) - \frac { 1 } { 90 } \sin ( 6 x )
سؤال
Solve the differential equation using the method of undetermined coefficients. y+6y+9y=2+xy ^ { \prime \prime } + 6 y ^ { \prime } + 9 y = 2 + x

A) y(x)=c1e3x+c2xe3x+29x+427y ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x } + \frac { 2 } { 9 } x + \frac { 4 } { 27 }
B) y(x)=c1e3x+c2xe3xy ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x }
C) y(x)=c1e3x+c2xe3x+29xy ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x } + \frac { 2 } { 9 } x
D) y(x)=c1e3x+29x+427y ( x ) = c _ { 1 } e ^ { - 3 x } + \frac { 2 } { 9 } x + \frac { 4 } { 27 }
E) y(x)=c1xe3x+c2x2e3x+427xy ( x ) = c _ { 1 } x e ^ { - 3 x } + c _ { 2 } x ^ { 2 } e ^ { - 3 x } + \frac { 4 } { 27 } x
سؤال
Solve the differential equation using the method of variation of parameters. y+y=secx,π4<x<π2y ^ { \prime \prime } + y = \sec x , \frac { \pi } { 4 } < x < \frac { \pi } { 2 }

A) y(x)=c1+[c2+ln(sinx)]sinxy ( x ) = c _ { 1 } + \left[ c _ { 2 } + \ln ( \sin x ) \right] \sin x
B) y(x)=(c1+x)sinx+c2+ln(cosx)y ( x ) = \left( c _ { 1 } + x \right) \sin x + c _ { 2 } + \ln ( \cos x )
C) y(x)=π2sinx+[π2+ln(cosx)]cosxy ( x ) = \frac { \pi } { 2 } \sin x + \left[ \frac { \pi } { 2 } + \ln ( \cos x ) \right] \cos x
D) y(x)=[π4+ln(cosx)]cosxy ( x ) = \left[ \frac { \pi } { 4 } + \ln ( \cos x ) \right] \cos x
E) y(x)=(c1+x)sinx+[c2+ln(cosx)]cosxy ( x ) = \left( c _ { 1 } + x \right) \sin x + \left[ c _ { 2 } + \ln ( \cos x ) \right] \cos x
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
سؤال
Find a trial solution for the method of undetermined coefficients. Do not determine the coefficients. y+2y=1+xe4xy ^ { \prime \prime } + 2 y ^ { \prime } = 1 + x e ^ { - 4 x }

A) yp1(x)=Ayp2(x)=Bxe4x\begin{array} { l } y _ { p 1 } ( x ) = A \\y _ { p 2 } ( x ) = B x e ^ { 4 x }\end{array}
B) yp1(x)=Axyp2(x)=x(Ax+B)e4x\begin{array} { l } y _ { p 1 } ( x ) = A x \\y _ { p 2 } ( x ) = x ( A x + B ) e ^ { - 4 x }\end{array}
C) y71(x)=Ayp2(x)=x(A+B)e4x\begin{array} { l } y _ { 71 } ( x ) = A \\y _ { p 2 } ( x ) = x ( A + B ) e ^ { - 4 x }\end{array}
D) yp1(x)=Axyp2(x)=x2(Ax+B)e4x\begin{array} { l } y _ { p 1 } ( x ) = A x \\y _ { p 2 } ( x ) = x ^ { 2 } ( A x + B ) e ^ { - 4 x }\end{array}
E) yp1(x)=Ayp2(x)=x(A+Bx)e4x\begin{array} { l } y _ { p 1 } ( x ) = A \\y _ { p 2 } ( x ) = x ( A + B x ) e ^ { - 4 x }\end{array}
سؤال
Solve the initial-value problem using the method of undetermined coefficients. Solve the initial-value problem using the method of undetermined coefficients.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of undetermined coefficients. y+5y+6y=x2y ^ { \prime \prime } + 5 y ^ { \prime } + 6 y = x ^ { 2 }

A) y(x)=c1e2x+c2e3xy ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x }
B) y(x)=16x2518x+19108y ( x ) = \frac { 1 } { 6 } x ^ { 2 } - \frac { 5 } { 18 } x + \frac { 19 } { 108 }
C) y(x)=c1e2x+c2e3x+16x2y ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x } + \frac { 1 } { 6 } x ^ { 2 }
D) y(x)=c1e2x+c2e3x+16x2518x+19108y ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x } + \frac { 1 } { 6 } x ^ { 2 } - \frac { 5 } { 18 } x + \frac { 19 } { 108 }
E) y(x)=c1e2x+c2e3x+9108c3y ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x } + \frac { 9 } { 108 } c _ { 3 }
سؤال
Graph the particular solution and several other solutions. 2y+3y+y=2+cos2x2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x

A)  <strong>Graph the particular solution and several other solutions.  2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x </strong> A)   B)   C)   <div style=padding-top: 35px>
B)  <strong>Graph the particular solution and several other solutions.  2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x </strong> A)   B)   C)   <div style=padding-top: 35px>
C)  <strong>Graph the particular solution and several other solutions.  2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x </strong> A)   B)   C)   <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of variation of parameters. y4y+3y=2sinxy ^ { \prime \prime } - 4 y ^ { \prime } + 3 y = 2 \sin x

A) y(x)=c1sinx+c25x+210sinxy ( x ) = c _ { 1 } \sin x + \frac { c _ { 2 } } { 5 } x + \frac { 2 } { 10 } \sin x
B) y(x)=c1e3x+c2ex+25sinxy ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 5 } \sin x
C) y(x)=c1sin3x+c24x+cos3xy ( x ) = c _ { 1 } \sin 3 x + \frac { c _ { 2 } } { 4 } x + \cos 3 x
D) y(x)=c1e3x+c2ex+210sinxy ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 10 } \sin x
E) y(x)=c1e3x+c2ex+25cosx+210sinxy ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 5 } \cos x + \frac { 2 } { 10 } \sin x
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
سؤال
Solve the initial-value problem using the method of undetermined coefficients. Solve the initial-value problem using the method of undetermined coefficients.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.  <div style=padding-top: 35px>
سؤال
Find a trial solution for the method of undetermined coefficients. Do not determine the coefficients. y+3y+5y=x4e9xy ^ { \prime \prime } + 3 y ^ { \prime } + 5 y = x ^ { 4 } e ^ { 9 x }

A) yy(x)=(Ax4+Bx2+C)e9xy _ { y } ( x ) = \left( A x ^ { 4 } + B x ^ { 2 } + C \right) e ^ { - 9 x }
B) yy(x)=(Ax3+Bx2+Cx+D)e9xy _ { y } ( x ) = \left( A x ^ { 3 } + B x ^ { 2 } + C x + D \right) e ^ { 9 x }
C) yy(x)=(Ax4+B)e9xy _ { y } ( x ) = \left( A x ^ { 4 } + B \right) e ^ { 9 x }
D) yy(x)=Ax9y _ { y } ( x ) = A x ^ { 9 }
E) yy(x)=(Ax4+Bx3+Cx2+Dx+E)e9xy _ { y } ( x ) = \left( A x ^ { 4 } + B x ^ { 3 } + C x ^ { 2 } + D x + E \right) e ^ { 9 x }
سؤال
Solve the differential equation using the method of undetermined coefficients. Solve the differential equation using the method of undetermined coefficients.  <div style=padding-top: 35px>
سؤال
Solve the differential equation using the method of variation of parameters. yy=e2xy ^ { \prime \prime } - y ^ { \prime \prime } = e ^ { 2 x }

A) y(x)=c1+c2ex+xe2x2y ( x ) = c _ { 1 } + c _ { 2 } e ^ { x } + \frac { x e ^ { 2 x } } { 2 }
B) y(x)=c1+c2xex+2e2xy ( x ) = c _ { 1 } + c _ { 2 } x e ^ { x } + 2 e ^ { 2 x }
C) y(x)=c1+xe2xy ( x ) = c _ { 1 } + x e ^ { 2 x }
D) y(x)=c1+c2xe2xy ( x ) = c _ { 1 } + c _ { 2 } x e ^ { 2 x }
E) y(x)=c1+(1+x)xe2xy ( x ) = c _ { 1 } + ( 1 + x ) x e ^ { 2 x }
سؤال
Solve the differential equation. y4y+13y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 13 y = 0

A) y(x)=ce2xcos(3x)y ( x ) = c e ^ { 2 x } \cos ( 3 x )
B) y(x)=e2x(c1cos(3x)+c2xsin(3x))y ( x ) = e ^ { 2 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x ) \right)
C) y(x)=e2x(c1cos(3x)+c2sin(3x))y ( x ) = e ^ { 2 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x ) \right)
D) y(x)=e3x(c1cos(2x)+c2sin(2x))y ( x ) = e ^ { 3 x } \left( c _ { 1 } \cos ( 2 x ) + c _ { 2 } \sin ( 2 x ) \right)
E) y(x)=e2x(c1cos(2x)+c2xsin(2x))y ( x ) = e ^ { 2 x } \left( c _ { 1 } \cos ( 2 x ) + c _ { 2 } x \sin ( 2 x ) \right)
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
سؤال
Solve the differential equation. y8y+25y=0y ^ { \prime \prime } - 8 y ^ { \prime } + 25 y = 0

A) y(x)=ce4xcos(3x)y ( x ) = c e ^ { 4 x } \cos ( 3 x )
B) y(x)=e4x(c1cos(3x)+c2xsin(3x))y ( x ) = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x ) \right)
C) y(x)=e4x(c1cos(4x)+c2xsin(4x))y ( x ) = e ^ { 4 x } \left( c _ { 1 } \cos ( 4 x ) + c _ { 2 } x \sin ( 4 x ) \right)
D) y(x)=e3x(c1cos(4x)+c2sin(4x))y ( x ) = e ^ { 3 x } \left( c _ { 1 } \cos ( 4 x ) + c _ { 2 } \sin ( 4 x ) \right)
E) y(x)=e4x(c1cos(3x)+c2sin(3x))y ( x ) = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x ) \right)
سؤال
Solve the boundary-value problem, if possible. Solve the boundary-value problem, if possible.  <div style=padding-top: 35px>
سؤال
Solve the differential equation. y8y+25y=0y ^ { \prime \prime } - 8 y ^ { \prime } + 25 y = 0

A) y=c1cos(3x)+c2xsin(3x)y = c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x )
B) y=ce4xcos(3x)y = c e ^ { 4 x } \cos ( 3 x )
C) y=e4x(c1cos(4x)+c2xsin(4x))y = e ^ { 4 x } \left( c _ { 1 } \cos ( 4 x ) + c _ { 2 } x \sin ( 4 x ) \right)
D) y=e4x(c1cos(3x)+c2sin(3x))y = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x ) \right)
E) y=e4x(c1cos(3x)+c2xsin(3x))y = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x ) \right)
سؤال
Solve the differential equation. 4y+y=04 y ^ { \prime \prime } + y = 0

A) y(x)=c1cos(x4)+c2sin(x2)y ( x ) = c _ { 1 } \cos \left( - \frac { x } { 4 } \right) + c _ { 2 } \sin \left( - \frac { x } { 2 } \right)
B) y(x)=c1cos(4x)c2sin(4x)y ( x ) = c _ { 1 } \cos ( 4 x ) - c _ { 2 } \sin ( 4 x )
C) y(x)=c1cos(2x)+c2sin(2x)y ( x ) = c _ { 1 } \cos ( 2 x ) + c _ { 2 } \sin ( 2 x )
D) y(x)=c1cos(2x)+c2sin(2x)y ( x ) = c _ { 1 } \cos ( - 2 x ) + c _ { 2 } \sin ( - 2 x )
E) y(x)=c1cos(x2)+c2sin(x2)y ( x ) = c _ { 1 } \cos \left( \frac { x } { 2 } \right) + c _ { 2 } \sin \left( \frac { x } { 2 } \right)
سؤال
Solve the boundary-value problem, if possible. y+5y36y=0,y(0)=0,y(2)=1y ^ { \prime \prime } + 5 y ^ { \prime } - 36 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1

A) y(x)=(e8e18)1(e4xe9x)y ( x ) = \left( e ^ { 8 } - e ^ { - 18 } \right) ^ { - 1 } \left( e ^ { 4 x } - e ^ { - 9 x } \right)
B) y(x)=(e4e18)(e4xe9x)y ( x ) = \left( e ^ { 4 } - e ^ { - 18 } \right) \left( e ^ { 4 x } - e ^ { - 9 x } \right)
C) y(x)=(e9e18)1(exe9x)y ( x ) = \left( e ^ { 9 } - e ^ { - 18 } \right) ^ { - 1 } \left( e ^ { x } - e ^ { - 9 x } \right)
D) y(x)=(e4e18)1(e4xe9x)y ( x ) = \left( e ^ { 4 } - e ^ { - 18 } \right) ^ { - 1 } \left( e ^ { 4 x } - e ^ { - 9 x } \right)
E) No solution
سؤال
Solve the initial-value problem. Solve the initial-value problem.  <div style=padding-top: 35px>
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
سؤال
Solve the differential equation. d2ydt2+dydt+3y=0\frac { d ^ { 2 } y } { d t ^ { 2 } } + \frac { d y } { d t } + 3 y = 0

A) y(t)=tet2[c1cos(11t2)+c2sin(11t2)]y ( t ) = t e ^ { \frac { t } { 2 } } \left[ c _ { 1 } \cos \left( \frac { \sqrt { 11 } t } { 2 } \right) + c _ { 2 } \sin \left( \frac { \sqrt { 11 } t } { 2 } \right) \right]
B) y(t)=e22[c1cos(t2)+c2sin(t2)]y ( t ) = e ^ { \frac { \sqrt { 2 } } { 2 } } \left[ c _ { 1 } \cos \left( \frac { \sqrt { t } } { 2 } \right) + c _ { 2 } \sin \left( \frac { \sqrt { t } } { 2 } \right) \right]
C) y(t)=et2[c1cos(11t2)+c2sin(11t2)]y ( t ) = e ^ { - \frac { t } { 2 } } \left[ c _ { 1 } \cos \left( \frac { \sqrt { 11 } t } { 2 } \right) + c _ { 2 } \sin \left( \frac { \sqrt { 11 } t } { 2 } \right) \right]
D) y(t)=t2et2[c1cos(11t2)+c2sin(11t2)]y ( t ) = t ^ { 2 } e ^ { - \frac { t } { 2 } } \left[ c _ { 1 } \cos \left( \sqrt { \frac { 11 t } { 2 } } \right) + c _ { 2 } \sin \left( \sqrt { \frac { 11 t } { 2 } } \right) \right]
E) y(t)=cet2cos(11t2)y ( t ) = c e ^ { - \frac { t } { 2 } } \cos \left( \sqrt { \frac { 11 t } { 2 } } \right)
سؤال
Solve the differential equation. y+5y6y=0y ^ { \prime \prime } + 5 y ^ { \prime } - 6 y = 0

A) y=C1ex+C2e7xy = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 7 x }
B) y=C1e6x+C2e6xy = C _ { 1 } e ^ { - 6 x } + C _ { 2 } e ^ { - 6 x }
C) y=C1ex3+C2e6x2y = \frac { C _ { 1 } e ^ { x } } { 3 } + \frac { C _ { 2 } e ^ { - 6 x } } { 2 }
D) y=C1ex+C2e6xy = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 6 x }
E) None of these
سؤال
Solve the initial-value problem. y2y24y=0,y(1)=4,y(1)=5y ^ { \prime \prime } - 2 y ^ { \prime } - 24 y = 0 , y ( 1 ) = 4 , y ^ { \prime } ( 1 ) = 5 .

A) y(x)=110e4(x1)+123e6(x1)y ( x ) = \frac { 1 } { 10 } e ^ { 4 ( x - 1 ) } + \frac { 1 } { 23 } e ^ { - 6 ( x - 1 ) }
B) y(x)=110e4x123e6xy ( x ) = \frac { 1 } { 10 } e ^ { 4 x } - \frac { 1 } { 23 } e ^ { - 6 x }
C) y(x)=e6xe4xy ( x ) = e ^ { 6 x } - e ^ { - 4 x }
D) y(x)=2110e6(x1)+1910e4(x1)y ( x ) = \frac { 21 } { 10 } e ^ { 6 ( x - 1 ) } + \frac { 19 } { 10 } e ^ { - 4 ( x - 1 ) }
E) y(x)=110ex123exy ( x ) = \frac { 1 } { 10 } e ^ { x } - \frac { 1 } { 23 } e ^ { - x }
سؤال
Solve the initial-value problem. Solve the initial-value problem.  <div style=padding-top: 35px>
سؤال
Find f by solving the initial value problem. f(x)=8x2+6x+4f ^ { \prime \prime } ( x ) = 8 x ^ { 2 } + 6 x + 4 ; f(1)=7f ( - 1 ) = - 7 , f(1)=3f ^ { \prime } ( - 1 ) = - 3

A) 83\frac { 8 } { 3 } x4x ^ { 4 } +3+ 3 x3x ^ { 3 } + 4 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x - 10
B) 23\frac { 2 } { 3 } x4x ^ { 4 } ++ x3x ^ { 3 } + 2 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x 8- 8
C) 23\frac { 2 } { 3 } x4x ^ { 4 } ++ x3x ^ { 3 } + 2 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x 16- 16
D) 83\frac { 8 } { 3 } x4x ^ { 4 } +3+ 3 x3x ^ { 3 } + 4 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x - 5
سؤال
Solve the boundary-value problem, if possible. y+16y+64y=0,y(0)=0,y(1)=4y ^ { \prime \prime } + 16 y ^ { \prime } + 64 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 4

A) y(x)=4xe(8x+8)y ( x ) = 4 x e ^ { ( - 8 x + 8 ) }
B) y(x)=4e(8x8)y ( x ) = 4 e ^ { ( - 8 x - 8 ) }
C) y(x)=4xe(8x8)y ( x ) = 4 x e ^ { ( - 8 x - 8 ) }
D) y(x)=8xe(4x+4)y ( x ) = 8 x e ^ { ( - 4 x + 4 ) }
E) No solution
سؤال
Solve the initial-value problem. y+8y+41y=0,y(0)=1,y(0)=7y ^ { \prime \prime } + 8 y ^ { \prime } + 41 y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 7

A) y=e5x(cos(5x)+115sin(5x))y = e ^ { 5 x } \left( \cos ( 5 x ) + \frac { 11 } { 5 } \sin ( 5 x ) \right)
B) y=e5x(cos(5x)+25sin(5x))y = e ^ { - 5 x } \left( \cos ( 5 x ) + \frac { 2 } { 5 } \sin ( 5 x ) \right)
C) y=e5x(cos(4x)+115sin(4x))y = e ^ { 5 x } \left( \cos ( 4 x ) + \frac { 11 } { 5 } \sin ( 4 x ) \right)
D) y=e4x(cos(5x)+115sin(5x))y = e ^ { - 4 x } \left( \cos ( 5 x ) + \frac { 11 } { 5 } \sin ( 5 x ) \right)
E) y=e4x(cos(5x)25sin(5x))y = e ^ { 4 x } \left( \cos ( 5 x ) - \frac { 2 } { 5 } \sin ( 5 x ) \right)
سؤال
Solve the initial-value problem. y+81y=0,y(π9)=0,y(π9)=8y ^ { \prime \prime } + 81 y = 0 , y \left( \frac { \pi } { 9 } \right) = 0 , y ^ { \prime } \left( \frac { \pi } { 9 } \right) = 8

A) y(x)=xsin(9x)y ( x ) = - x \sin ( 9 x )
B) y(x)=89xsin(9x)y ( x ) = \frac { 8 } { 9 } x \sin ( 9 x )
C) y(x)=89xsin(9x)y ( x ) = - \frac { 8 } { 9 } x \sin ( 9 x )
D) y(x)=89sin(9x)y ( x ) = - \frac { 8 } { 9 } \sin ( 9 x )
E) y(x)=89sin(9x)y ( x ) = \frac { 8 } { 9 } \sin ( 9 x )
سؤال
Solve the initial value problem. Solve the initial value problem.  <div style=padding-top: 35px>
سؤال
Solve the initial-value problem. y+2y3y=0,y(0)=2,y(0)=1y ^ { \prime \prime } + 2 y ^ { \prime } - 3 y = 0 , y ( 0 ) = 2 , y ^ { \prime } ( 0 ) = 1

A) y=14e2x+74e2xy = \frac { 1 } { 4 } e ^ { 2 x } + \frac { 7 } { 4 } e ^ { - 2 x }
B) y=ex+14e3xy = e ^ { x } + \frac { 1 } { 4 } e ^ { - 3 x }
C) y=e2x+14e2xy = e ^ { 2 x } + \frac { 1 } { 4 } e ^ { - 2 x }
D) y=74ex+14e3xy = \frac { 7 } { 4 } e ^ { x } + \frac { 1 } { 4 } e ^ { - 3 x }
E) None of these
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
سؤال
Solve the boundary-value problem, if possible. Solve the boundary-value problem, if possible.  <div style=padding-top: 35px>
سؤال
Solve the initial-value problem. Solve the initial-value problem.  <div style=padding-top: 35px>
سؤال
Solve the differential equation. Solve the differential equation.  <div style=padding-top: 35px>
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Deck 17: Second-Order Differential Equations
1
A spring with a mass of 2 kg has damping constant 14, and a force of A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. The spring is stretched 1m beyond its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t. N is required to keep the spring stretched A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. The spring is stretched 1m beyond its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t. m beyond its natural length. The spring is stretched 1m beyond its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t.
2
Solve the differential equation using the method of variation of parameters. y+4y+4y=e2xx3y ^ { \prime \prime } + 4 y ^ { \prime } + 4 y = \frac { e ^ { - 2 x } } { x ^ { 3 } }

A) y(x)=e2x(c1+c212x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } \frac { 1 } { 2 x } \right)
B) y(x)=e2x(c1+c2x+12x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x + \frac { 1 } { 2 x } \right)
C) y(x)=e2x(c1+c2)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } \right)
D) y(x)=e2x(c1+c2x)+12xy ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x \right) + \frac { 1 } { 2 x }
E) y(x)=e2x(c1+c2x+c312x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x + c _ { 3 } \frac { 1 } { 2 x } \right)
y(x)=e2x(c1+c2x+12x)y ( x ) = e ^ { - 2 x } \left( c _ { 1 } + c _ { 2 } x + \frac { 1 } { 2 x } \right)
3
Use power series to solve the differential equation.. (x2+1)y+xyy=0\left( x ^ { 2 } + 1 \right) y ^ { \prime \prime } + x y ^ { \prime } - y = 0

A) y(x)=c0n=2(1)n1(2n3)!22n2n!(n2)!x2ny ( x ) = c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n }
B) y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2n!(n2)!x2n+1y ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n + 1 }
C) y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2(n2)!x2ny ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } ( n - 2 ) ! } x ^ { 2 n }
D) y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2n!(n2)!x2ny ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n }
E) y(x)=c0+c1x+c0x22+c0n=2(1)n+1(2n)!22n+2n!(n2)!x2n+1y ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } ( 2 n ) ! } { 2 ^ { 2 n + 2 } n ! ( n - 2 ) ! } x ^ { 2 n + 1 }
y(x)=c0+c1x+c0x22+c0n=2(1)n1(2n3)!22n2n!(n2)!x2ny ( x ) = c _ { 0 + } c _ { 1 } x + c _ { 0 } \frac { x ^ { 2 } } { 2 } + c _ { 0 } \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } ( 2 n - 3 ) ! } { 2 ^ { 2 n - 2 } n ! ( n - 2 ) ! } x ^ { 2 n }
4
Use power series to solve the differential equation. y=25yy ^ { \prime \prime } = 25 y

A) y(x)=c1cosh5x+c2sinh5xy ( x ) = c _ { 1 } \cosh 5 x + c _ { 2 } \sinh 5 x
B) y(x)=c15coshx+c25sinhx+exy ( x ) = c _ { 1 } 5 \cosh x + c _ { 2 } 5 \sinh x + e ^ { x }
C) y(x)=5coshx+5sinhxy ( x ) = 5 \cosh x + 5 \sinh x
D) y(x)=c1cosh5x+c2sinh5x+exy ( x ) = c _ { 1 } \cosh 5 x + c _ { 2 } \sinh 5 x + e ^ { x }
E) y(x)=c1cosh5x+c2sinh5x+x2y ( x ) = c _ { 1 } \cosh 5 x + c _ { 2 } \sinh 5 x + x ^ { 2 }
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5
A series circuit consists of a resistor R=32ΩR = 32 \Omega , an inductor with L=4HL = 4 H , a capacitor with C=0.003125 FC = 0.003125 \mathrm {~F} , and a 2424 -V battery. If the initial charge is 0.0008 C and the initial current is 0, find the current I(t) at time t.

A) I(t)=0.742e4tsin(8t)I ( t ) = 0.742 e ^ { - 4 t } \sin ( 8 t )
B) I(t)=e4t160cos(8t)I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t )
C) I(t)=e4t160cos(8t)12I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t ) - \frac { 1 } { 2 }
D) I(t)=0.742e8tI ( t ) = 0.742 e ^ { - 8 t }
E) I(t)=e4t160(0.003125cos(4t)+8sin(4t))12I ( t ) = \frac { e ^ { - 4 t } } { 160 } ( 0.003125 \cos ( 4 t ) + 8 \sin ( 4 t ) ) - \frac { 1 } { 2 }
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6
Use power series to solve the differential equation. y=7x3yy ^ { \prime } = 7 x ^ { 3 } y

A) y(x)=7ex44y ( x ) = 7 e ^ { \frac { x ^ { 4 } } { 4 } }
B) y(x)=ce7xy ( x ) = c e ^ { 7 x }
C) y(x)=ex44y ( x ) = e ^ { - \frac { x ^ { 4 } } { 4 } }
D) y(x)=ce7x44y ( x ) = c e ^ { - \frac { 7 x ^ { 4 } } { 4 } }
E) y(x)=ce7x44y ( x ) = c e ^ { \frac { 7 x ^ { 4 } } { 4 } }
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7
A series circuit consists of a resistor A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t. an inductor with L = A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t. H, a capacitor with
C = A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t. F, and a A series circuit consists of a resistor   an inductor with L =   H, a capacitor with C =   F, and a   -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t. -V battery. If the initial charge and current are both 0, find the charge Q(t) at time t.
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8
A spring has a mass of 11 kg and its damping constant is c=10c = 10 . The spring starts from its equilibrium position with a velocity of 11 m/s. Graph the position function for the spring constant k=20k = 20 .

A)  <strong>A spring has a mass of  1  kg and its damping constant is  c = 10  . The spring starts from its equilibrium position with a velocity of  1  m/s. Graph the position function for the spring constant  k = 20  .</strong> A)   B)   C)
B)  <strong>A spring has a mass of  1  kg and its damping constant is  c = 10  . The spring starts from its equilibrium position with a velocity of  1  m/s. Graph the position function for the spring constant  k = 20  .</strong> A)   B)   C)
C)  <strong>A spring has a mass of  1  kg and its damping constant is  c = 10  . The spring starts from its equilibrium position with a velocity of  1  m/s. Graph the position function for the spring constant  k = 20  .</strong> A)   B)   C)
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9
Use power series to solve the differential equation. (x10)y+2y=0( x - 10 ) y ^ { \prime } + 2 y = 0

A) y(x)=c0n=0xn10ny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 10 ^ { n } }
B) y(x)=c0n=0n+110nxny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { n + 1 } { 10 ^ { n } } x ^ { n }
C) y(x)=c0n=0(n+1)xny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { ( n + 1 ) } { x ^ { n } }
D) y(x)=c0n=0(n+1)xny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } ( n + 1 ) x ^ { n }
E) y(x)=c0n=0nxn10ny ( x ) = c _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { n x ^ { n } } { 10 ^ { n } }
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10
A spring with a 1616 -kg mass has natural length 0.80.8 m and is maintained stretched to a length of 1.21.2 m by a force of 19.619.6 N. If the spring is compressed to a length of 0.40.4 m and then released with zero velocity, find the position x(t)x ( t ) of the mass at any time tt .

A) x(t)=0.4e4tcos(0.4t)x ( t ) = - 0.4 e ^ { - 4 t } \cos ( 0.4 t )
B) x(t)=0.8sin(1.75t)x ( t ) = 0.8 \sin ( 1.75 t )
C) x(t)=0.4e4tcos(1.75t)x ( t ) = - 0.4 e ^ { - 4 t } \cos ( 1.75 t )
D) x(t)=0.4cos(1.75t)+0.4sin(1.75t)x ( t ) = - 0.4 \cos ( 1.75 t ) + 0.4 \sin ( 1.75 t )
E) x(t)=0.4cos(1.75t)x ( t ) = - 0.4 \cos ( 1.75 t )
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11
Suppose a spring has mass M and spring constant k and let ω=k/M\omega = \sqrt { k / M } . Suppose that the damping constant is so small that the damping force is negligible. If an external force F(t)=8F0cos(ωt)F ( t ) = 8 F _ { 0 } \cos ( \omega t ) is applied (the applied frequency equals the natural frequency), use the method of undetermined coefficients to find the equation that describes the motion of the mass.

A) x(t)=c1cos(ωt)+c2sin(ωt)+F0t22Mωsin(ωt)x ( t ) = c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) + \frac { F _ { 0 } t ^ { - 2 } } { 2 M \omega } \sin ( \omega t )
B) x(t)=c1cos(ωt)+c2sin(ωt)+F0eωt2Mωx ( t ) = c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) + \frac { F _ { 0 } e ^ { - \omega t } } { 2 M \omega }
C) x(t)=c1cos(ωt)+c2sin(ωt)+4F0tMωsin(ωt)x ( t ) = c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) + \frac { 4 F _ { 0 } t } { M \omega } \sin ( \omega t )
D) x(t)=F0t22Mωcos(ωt)x ( t ) = \frac { F _ { 0 } t ^ { 2 } } { 2 M \omega } \cos ( \omega t )
E) x(t)=F0t2Mω(c1cos(ωt)+c2sin(ωt))x ( t ) = \frac { F _ { 0 } t } { 2 M \omega } \left( c _ { 1 } \cos ( \omega t ) + c _ { 2 } \sin ( \omega t ) \right)
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12
A spring with a mass of 2 kg has damping constant 14, and a force of A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. Find the mass that would produce critical damping. N is required to keep the spring stretched A spring with a mass of 2 kg has damping constant 14, and a force of   N is required to keep the spring stretched   m beyond its natural length. Find the mass that would produce critical damping. m beyond its natural length. Find the mass that would produce critical damping.
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13
A spring with a mass of 99 kg has damping constant 28 and spring constant 195195 . Find the damping constant that would produce critical damping.

A) c=c = 3619536 \sqrt { 195 }
B) c=c = 23402340
C) c=1959c = 195 \sqrt { 9 }
D) c=c = 9 195\sqrt { 195 }
E) c=c = 61956 \sqrt { 195 }
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14
Use power series to solve the differential equation. yxyy=0,y(0)=7,y(0)=0y ^ { \prime \prime } - x y ^ { \prime } - y = 0 , y ( 0 ) = 7 , y ^ { \prime } ( 0 ) = 0

A) y(x)=7exy ( x ) = 7 e ^ { x }
B) y(x)=7ex22y ( x ) = 7 e ^ { - \frac { x ^ { 2 } } { 2 } }
C) y(x)=e7xy ( x ) = e ^ { 7 x }
D) y(x)=cex22y ( x ) = c e ^ { \frac { x ^ { 2 } } { 2 } }
E) y(x)=7ex22y ( x ) = 7 e ^ { \frac { x ^ { 2 } } { 2 } }
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15
A spring with a 3-kg mass is held stretched 0.9 m beyond its natural length by a force of 30 N. If the spring begins at its equilibrium position but a push gives it an initial velocity of 11 m/s, find the position x(t) of the mass after t seconds.

A) x(t)=1sin(7t)x ( t ) = 1 \sin ( 7 t )
B) x(t)=0.3sin(103t)+0.4cos(103t)x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right) + 0.4 \cos \left( \frac { 10 } { 3 } t \right)
C) x(t)=0.3sin(103t)x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right)
D) x(t)=0.4cos(103t)x ( t ) = 0.4 \cos \left( \frac { 10 } { 3 } t \right)
E) x(t)=103sin(7t)x ( t ) = \frac { 10 } { 3 } \sin ( 7 t )
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16
The solution of the initial-value problem x2y+xy+x2y=0,y(0)=1,y(0)=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is called a Bessel function of order 0. Solve the initial - value problem to find a power series expansion for the Bessel function.

A) y(x)=n=0(1)n+1nx2n122n1(n!)2y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { n x ^ { 2 n - 1 } } { 2 ^ { 2 n - 1 } ( n ! ) ^ { 2 } }
B) y(x)=n=0(1)nxn2n(2n!)y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { n } } { 2 ^ { n } ( 2 n ! ) }
C) y(x)=n=0(1)nx2n22n(n!)2y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { 2 ^ { 2 n } ( n ! ) ^ { 2 } }
D) y(x)=2x+n=0(1)nx2n(n!)2y ( x ) = 2 ^ { x } + \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { ( n ! ) ^ { 2 } }
E) y(x)=x+n=0(1)nx2n22n(n!)2y ( x ) = x + \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { 2 ^ { 2 n } ( n ! ) ^ { 2 } }
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17
A series circuit consists of a resistor R=96ΩR = 96 \Omega , an inductor with L=8HL = 8 \mathrm { H } , a capacitor with C=0.00125 FC = 0.00125 \mathrm {~F} , and a generator producing a voltage of E(t)=48cos(10t)E ( t ) = 48 \cos ( 10 t ) If the initial charge is Q=0.001CQ = 0.001 \mathrm { C } and the initial current is 0, find the charge Q(t)Q ( t ) at time t.

A) Q(t)=e6t(0.001cos(8t)0.06175sin(8t))+120sin(10t)Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 8 t ) - 0.06175 \sin ( 8 t ) ) + \frac { 1 } { 20 } \sin ( 10 t )
B) Q(t)=e2t(0.051cos(2t)0.06175sin(2t))+120sin(10t)Q ( t ) = e ^ { - 2 t } ( 0.051 \cos ( 2 t ) - 0.06175 \sin ( 2 t ) ) + \frac { 1 } { 20 } \sin ( 10 t )
C) Q(t)=e6t(cos(6t)0.06175sin(6t))Q ( t ) = e ^ { - 6 t } ( \cos ( 6 t ) - 0.06175 \sin ( 6 t ) )
D) Q(t)=e3t(cos(5t)0.06175sin(5t))Q ( t ) = e ^ { - 3 t } ( \cos ( 5 t ) - 0.06175 \sin ( 5 t ) )
E) Q(t)=e6t(0.001cos(8t)0.06175sin(8t))Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 8 t ) - 0.06175 \sin ( 8 t ) )
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18
A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.

A) <strong>A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.</strong> A)   B)   C)
B) <strong>A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.</strong> A)   B)   C)
C) <strong>A spring with a mass of 2 kg has damping constant 8 and spring constant 80. Graph the position function of the mass at time t if it starts at the equilibrium position with a velocity of 2 m/s.</strong> A)   B)   C)
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19
The figure shows a pendulum with length L and the angle θ\theta from the vertical to the pendulum. It can be shown that θ\theta , as a function of time, satisfies the nonlinear differential equation d2θdt2+gLsinθ=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0 where g=9.8 m/s2g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }  is the acceleration due to gravity. For small values of \text { is the acceleration due to gravity. For small values of } θ\theta we can use the linear approximation sinθ=θ\sin \theta = \theta  and then the differential equation becomes linear. Find the equation \text { and then the differential equation becomes linear. Find the equation }  of motion of a pendulum with length 1 m if θ is initially 0.2rad and the initial \text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }  angular velocity is \text { angular velocity is } dθdt=1rad/s\frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }  <strong>The figure shows a pendulum with length L and the angle  \theta  from the vertical to the pendulum. It can be shown that  \theta  , as a function of time, satisfies the nonlinear differential equation  \frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0  where  g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }   \text { is the acceleration due to gravity. For small values of }   \theta  we can use the linear approximation  \sin \theta = \theta   \text { and then the differential equation becomes linear. Find the equation }   \text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }   \text { angular velocity is }   \frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }   </strong> A)  \theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )  B)  \theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )  C)  \theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )  D)  \theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )  E)  \theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )

A) θ(t)=0.2cos(9.8t)+19.8sin(9.8t)\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )
B) θ(t)=0.2cos(9.8t)+2sin(9.8t)\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )
C) θ(t)=2cos(9.8t)+19.8sin(9.8t)\theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )
D) θ(t)=19.8cos(9.8t)+0.2sin(9.8t)\theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )
E) θ(t)=0.2sin(9.8t)+19.8cos(9.8t)\theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )
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20
Use power series to solve the differential equation. y+x2y=0,y(0)=6,y(0)=0y ^ { \prime \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { \prime } ( 0 ) = 0

A) y(x)=6n=1(1)nx4n4n(4n1)(4n4)(4n5)43y ( x ) = 6 - \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
B) y(x)=n=1(1)n6x4n4n(4n1)(4n4)(4n5)43y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
C) y(x)=6+n=1(1)n6x4n4n(4n1)(4n4)(4n5)43y ( x ) = 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
D) y(x)=6+n=1(1)nx4n4n(4n1)(4n4)(4n5)43y ( x ) = - 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
E) y(x)=n=1(1)nx4n4n(4n1)(4n4)(4n5)43y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
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21
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.
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22
Solve the differential equation using the method of undetermined coefficients. Solve the differential equation using the method of undetermined coefficients.
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23
Solve the differential equation using the method of undetermined coefficients. y3y=sin9xy ^ { \prime \prime } - 3 y ^ { \prime } = \sin 9 x

A) y(x)=c1e3x+1270cos(6x)190sin(6x)y ( x ) = c _ { 1 } e ^ { 3 x } + \frac { 1 } { 270 } \cos ( 6 x ) - \frac { 1 } { 90 } \sin ( 6 x )
B) y(x)=c1+c2e3x+1270cos(6x)190sin(6x)y ( x ) = c _ { 1 } + c _ { 2 } e ^ { 3 x } + \frac { 1 } { 270 } \cos ( 6 x ) - \frac { 1 } { 90 } \sin ( 6 x )
C) y(x)=c1e3x+190cos(6x)+1270sin(6x)y ( x ) = c _ { 1 } e ^ { 3 x } + \frac { 1 } { 90 } \cos ( 6 x ) + \frac { 1 } { 270 } \sin ( 6 x )
D) y(x)=c1+c2e3x190sin(6x)y ( x ) = c _ { 1 } + c _ { 2 } e ^ { 3 x } - \frac { 1 } { 90 } \sin ( 6 x )
E) y(x)=1270cos(6x)190sin(6x)y ( x ) = \frac { 1 } { 270 } \cos ( 6 x ) - \frac { 1 } { 90 } \sin ( 6 x )
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24
Solve the differential equation using the method of undetermined coefficients. y+6y+9y=2+xy ^ { \prime \prime } + 6 y ^ { \prime } + 9 y = 2 + x

A) y(x)=c1e3x+c2xe3x+29x+427y ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x } + \frac { 2 } { 9 } x + \frac { 4 } { 27 }
B) y(x)=c1e3x+c2xe3xy ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x }
C) y(x)=c1e3x+c2xe3x+29xy ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x } + \frac { 2 } { 9 } x
D) y(x)=c1e3x+29x+427y ( x ) = c _ { 1 } e ^ { - 3 x } + \frac { 2 } { 9 } x + \frac { 4 } { 27 }
E) y(x)=c1xe3x+c2x2e3x+427xy ( x ) = c _ { 1 } x e ^ { - 3 x } + c _ { 2 } x ^ { 2 } e ^ { - 3 x } + \frac { 4 } { 27 } x
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25
Solve the differential equation using the method of variation of parameters. y+y=secx,π4<x<π2y ^ { \prime \prime } + y = \sec x , \frac { \pi } { 4 } < x < \frac { \pi } { 2 }

A) y(x)=c1+[c2+ln(sinx)]sinxy ( x ) = c _ { 1 } + \left[ c _ { 2 } + \ln ( \sin x ) \right] \sin x
B) y(x)=(c1+x)sinx+c2+ln(cosx)y ( x ) = \left( c _ { 1 } + x \right) \sin x + c _ { 2 } + \ln ( \cos x )
C) y(x)=π2sinx+[π2+ln(cosx)]cosxy ( x ) = \frac { \pi } { 2 } \sin x + \left[ \frac { \pi } { 2 } + \ln ( \cos x ) \right] \cos x
D) y(x)=[π4+ln(cosx)]cosxy ( x ) = \left[ \frac { \pi } { 4 } + \ln ( \cos x ) \right] \cos x
E) y(x)=(c1+x)sinx+[c2+ln(cosx)]cosxy ( x ) = \left( c _ { 1 } + x \right) \sin x + \left[ c _ { 2 } + \ln ( \cos x ) \right] \cos x
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26
Solve the differential equation. Solve the differential equation.
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27
Find a trial solution for the method of undetermined coefficients. Do not determine the coefficients. y+2y=1+xe4xy ^ { \prime \prime } + 2 y ^ { \prime } = 1 + x e ^ { - 4 x }

A) yp1(x)=Ayp2(x)=Bxe4x\begin{array} { l } y _ { p 1 } ( x ) = A \\y _ { p 2 } ( x ) = B x e ^ { 4 x }\end{array}
B) yp1(x)=Axyp2(x)=x(Ax+B)e4x\begin{array} { l } y _ { p 1 } ( x ) = A x \\y _ { p 2 } ( x ) = x ( A x + B ) e ^ { - 4 x }\end{array}
C) y71(x)=Ayp2(x)=x(A+B)e4x\begin{array} { l } y _ { 71 } ( x ) = A \\y _ { p 2 } ( x ) = x ( A + B ) e ^ { - 4 x }\end{array}
D) yp1(x)=Axyp2(x)=x2(Ax+B)e4x\begin{array} { l } y _ { p 1 } ( x ) = A x \\y _ { p 2 } ( x ) = x ^ { 2 } ( A x + B ) e ^ { - 4 x }\end{array}
E) yp1(x)=Ayp2(x)=x(A+Bx)e4x\begin{array} { l } y _ { p 1 } ( x ) = A \\y _ { p 2 } ( x ) = x ( A + B x ) e ^ { - 4 x }\end{array}
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28
Solve the initial-value problem using the method of undetermined coefficients. Solve the initial-value problem using the method of undetermined coefficients.
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29
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.
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30
Solve the differential equation using the method of undetermined coefficients. y+5y+6y=x2y ^ { \prime \prime } + 5 y ^ { \prime } + 6 y = x ^ { 2 }

A) y(x)=c1e2x+c2e3xy ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x }
B) y(x)=16x2518x+19108y ( x ) = \frac { 1 } { 6 } x ^ { 2 } - \frac { 5 } { 18 } x + \frac { 19 } { 108 }
C) y(x)=c1e2x+c2e3x+16x2y ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x } + \frac { 1 } { 6 } x ^ { 2 }
D) y(x)=c1e2x+c2e3x+16x2518x+19108y ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x } + \frac { 1 } { 6 } x ^ { 2 } - \frac { 5 } { 18 } x + \frac { 19 } { 108 }
E) y(x)=c1e2x+c2e3x+9108c3y ( x ) = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x } + \frac { 9 } { 108 } c _ { 3 }
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31
Graph the particular solution and several other solutions. 2y+3y+y=2+cos2x2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x

A)  <strong>Graph the particular solution and several other solutions.  2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x </strong> A)   B)   C)
B)  <strong>Graph the particular solution and several other solutions.  2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x </strong> A)   B)   C)
C)  <strong>Graph the particular solution and several other solutions.  2 y ^ { \prime \prime } + 3 y ^ { \prime } + y = 2 + \cos 2 x </strong> A)   B)   C)
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32
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.
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33
Solve the differential equation using the method of variation of parameters. y4y+3y=2sinxy ^ { \prime \prime } - 4 y ^ { \prime } + 3 y = 2 \sin x

A) y(x)=c1sinx+c25x+210sinxy ( x ) = c _ { 1 } \sin x + \frac { c _ { 2 } } { 5 } x + \frac { 2 } { 10 } \sin x
B) y(x)=c1e3x+c2ex+25sinxy ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 5 } \sin x
C) y(x)=c1sin3x+c24x+cos3xy ( x ) = c _ { 1 } \sin 3 x + \frac { c _ { 2 } } { 4 } x + \cos 3 x
D) y(x)=c1e3x+c2ex+210sinxy ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 10 } \sin x
E) y(x)=c1e3x+c2ex+25cosx+210sinxy ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 5 } \cos x + \frac { 2 } { 10 } \sin x
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34
Solve the differential equation. Solve the differential equation.
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35
Solve the initial-value problem using the method of undetermined coefficients. Solve the initial-value problem using the method of undetermined coefficients.
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36
Solve the differential equation using the method of variation of parameters. Solve the differential equation using the method of variation of parameters.
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37
Find a trial solution for the method of undetermined coefficients. Do not determine the coefficients. y+3y+5y=x4e9xy ^ { \prime \prime } + 3 y ^ { \prime } + 5 y = x ^ { 4 } e ^ { 9 x }

A) yy(x)=(Ax4+Bx2+C)e9xy _ { y } ( x ) = \left( A x ^ { 4 } + B x ^ { 2 } + C \right) e ^ { - 9 x }
B) yy(x)=(Ax3+Bx2+Cx+D)e9xy _ { y } ( x ) = \left( A x ^ { 3 } + B x ^ { 2 } + C x + D \right) e ^ { 9 x }
C) yy(x)=(Ax4+B)e9xy _ { y } ( x ) = \left( A x ^ { 4 } + B \right) e ^ { 9 x }
D) yy(x)=Ax9y _ { y } ( x ) = A x ^ { 9 }
E) yy(x)=(Ax4+Bx3+Cx2+Dx+E)e9xy _ { y } ( x ) = \left( A x ^ { 4 } + B x ^ { 3 } + C x ^ { 2 } + D x + E \right) e ^ { 9 x }
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38
Solve the differential equation using the method of undetermined coefficients. Solve the differential equation using the method of undetermined coefficients.
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39
Solve the differential equation using the method of variation of parameters. yy=e2xy ^ { \prime \prime } - y ^ { \prime \prime } = e ^ { 2 x }

A) y(x)=c1+c2ex+xe2x2y ( x ) = c _ { 1 } + c _ { 2 } e ^ { x } + \frac { x e ^ { 2 x } } { 2 }
B) y(x)=c1+c2xex+2e2xy ( x ) = c _ { 1 } + c _ { 2 } x e ^ { x } + 2 e ^ { 2 x }
C) y(x)=c1+xe2xy ( x ) = c _ { 1 } + x e ^ { 2 x }
D) y(x)=c1+c2xe2xy ( x ) = c _ { 1 } + c _ { 2 } x e ^ { 2 x }
E) y(x)=c1+(1+x)xe2xy ( x ) = c _ { 1 } + ( 1 + x ) x e ^ { 2 x }
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40
Solve the differential equation. y4y+13y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 13 y = 0

A) y(x)=ce2xcos(3x)y ( x ) = c e ^ { 2 x } \cos ( 3 x )
B) y(x)=e2x(c1cos(3x)+c2xsin(3x))y ( x ) = e ^ { 2 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x ) \right)
C) y(x)=e2x(c1cos(3x)+c2sin(3x))y ( x ) = e ^ { 2 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x ) \right)
D) y(x)=e3x(c1cos(2x)+c2sin(2x))y ( x ) = e ^ { 3 x } \left( c _ { 1 } \cos ( 2 x ) + c _ { 2 } \sin ( 2 x ) \right)
E) y(x)=e2x(c1cos(2x)+c2xsin(2x))y ( x ) = e ^ { 2 x } \left( c _ { 1 } \cos ( 2 x ) + c _ { 2 } x \sin ( 2 x ) \right)
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41
Solve the differential equation. Solve the differential equation.
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42
Solve the differential equation. y8y+25y=0y ^ { \prime \prime } - 8 y ^ { \prime } + 25 y = 0

A) y(x)=ce4xcos(3x)y ( x ) = c e ^ { 4 x } \cos ( 3 x )
B) y(x)=e4x(c1cos(3x)+c2xsin(3x))y ( x ) = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x ) \right)
C) y(x)=e4x(c1cos(4x)+c2xsin(4x))y ( x ) = e ^ { 4 x } \left( c _ { 1 } \cos ( 4 x ) + c _ { 2 } x \sin ( 4 x ) \right)
D) y(x)=e3x(c1cos(4x)+c2sin(4x))y ( x ) = e ^ { 3 x } \left( c _ { 1 } \cos ( 4 x ) + c _ { 2 } \sin ( 4 x ) \right)
E) y(x)=e4x(c1cos(3x)+c2sin(3x))y ( x ) = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x ) \right)
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43
Solve the boundary-value problem, if possible. Solve the boundary-value problem, if possible.
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44
Solve the differential equation. y8y+25y=0y ^ { \prime \prime } - 8 y ^ { \prime } + 25 y = 0

A) y=c1cos(3x)+c2xsin(3x)y = c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x )
B) y=ce4xcos(3x)y = c e ^ { 4 x } \cos ( 3 x )
C) y=e4x(c1cos(4x)+c2xsin(4x))y = e ^ { 4 x } \left( c _ { 1 } \cos ( 4 x ) + c _ { 2 } x \sin ( 4 x ) \right)
D) y=e4x(c1cos(3x)+c2sin(3x))y = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x ) \right)
E) y=e4x(c1cos(3x)+c2xsin(3x))y = e ^ { 4 x } \left( c _ { 1 } \cos ( 3 x ) + c _ { 2 } x \sin ( 3 x ) \right)
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45
Solve the differential equation. 4y+y=04 y ^ { \prime \prime } + y = 0

A) y(x)=c1cos(x4)+c2sin(x2)y ( x ) = c _ { 1 } \cos \left( - \frac { x } { 4 } \right) + c _ { 2 } \sin \left( - \frac { x } { 2 } \right)
B) y(x)=c1cos(4x)c2sin(4x)y ( x ) = c _ { 1 } \cos ( 4 x ) - c _ { 2 } \sin ( 4 x )
C) y(x)=c1cos(2x)+c2sin(2x)y ( x ) = c _ { 1 } \cos ( 2 x ) + c _ { 2 } \sin ( 2 x )
D) y(x)=c1cos(2x)+c2sin(2x)y ( x ) = c _ { 1 } \cos ( - 2 x ) + c _ { 2 } \sin ( - 2 x )
E) y(x)=c1cos(x2)+c2sin(x2)y ( x ) = c _ { 1 } \cos \left( \frac { x } { 2 } \right) + c _ { 2 } \sin \left( \frac { x } { 2 } \right)
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46
Solve the boundary-value problem, if possible. y+5y36y=0,y(0)=0,y(2)=1y ^ { \prime \prime } + 5 y ^ { \prime } - 36 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1

A) y(x)=(e8e18)1(e4xe9x)y ( x ) = \left( e ^ { 8 } - e ^ { - 18 } \right) ^ { - 1 } \left( e ^ { 4 x } - e ^ { - 9 x } \right)
B) y(x)=(e4e18)(e4xe9x)y ( x ) = \left( e ^ { 4 } - e ^ { - 18 } \right) \left( e ^ { 4 x } - e ^ { - 9 x } \right)
C) y(x)=(e9e18)1(exe9x)y ( x ) = \left( e ^ { 9 } - e ^ { - 18 } \right) ^ { - 1 } \left( e ^ { x } - e ^ { - 9 x } \right)
D) y(x)=(e4e18)1(e4xe9x)y ( x ) = \left( e ^ { 4 } - e ^ { - 18 } \right) ^ { - 1 } \left( e ^ { 4 x } - e ^ { - 9 x } \right)
E) No solution
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47
Solve the initial-value problem. Solve the initial-value problem.
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48
Solve the differential equation. Solve the differential equation.
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49
Solve the differential equation. Solve the differential equation.
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50
Solve the differential equation. d2ydt2+dydt+3y=0\frac { d ^ { 2 } y } { d t ^ { 2 } } + \frac { d y } { d t } + 3 y = 0

A) y(t)=tet2[c1cos(11t2)+c2sin(11t2)]y ( t ) = t e ^ { \frac { t } { 2 } } \left[ c _ { 1 } \cos \left( \frac { \sqrt { 11 } t } { 2 } \right) + c _ { 2 } \sin \left( \frac { \sqrt { 11 } t } { 2 } \right) \right]
B) y(t)=e22[c1cos(t2)+c2sin(t2)]y ( t ) = e ^ { \frac { \sqrt { 2 } } { 2 } } \left[ c _ { 1 } \cos \left( \frac { \sqrt { t } } { 2 } \right) + c _ { 2 } \sin \left( \frac { \sqrt { t } } { 2 } \right) \right]
C) y(t)=et2[c1cos(11t2)+c2sin(11t2)]y ( t ) = e ^ { - \frac { t } { 2 } } \left[ c _ { 1 } \cos \left( \frac { \sqrt { 11 } t } { 2 } \right) + c _ { 2 } \sin \left( \frac { \sqrt { 11 } t } { 2 } \right) \right]
D) y(t)=t2et2[c1cos(11t2)+c2sin(11t2)]y ( t ) = t ^ { 2 } e ^ { - \frac { t } { 2 } } \left[ c _ { 1 } \cos \left( \sqrt { \frac { 11 t } { 2 } } \right) + c _ { 2 } \sin \left( \sqrt { \frac { 11 t } { 2 } } \right) \right]
E) y(t)=cet2cos(11t2)y ( t ) = c e ^ { - \frac { t } { 2 } } \cos \left( \sqrt { \frac { 11 t } { 2 } } \right)
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51
Solve the differential equation. y+5y6y=0y ^ { \prime \prime } + 5 y ^ { \prime } - 6 y = 0

A) y=C1ex+C2e7xy = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 7 x }
B) y=C1e6x+C2e6xy = C _ { 1 } e ^ { - 6 x } + C _ { 2 } e ^ { - 6 x }
C) y=C1ex3+C2e6x2y = \frac { C _ { 1 } e ^ { x } } { 3 } + \frac { C _ { 2 } e ^ { - 6 x } } { 2 }
D) y=C1ex+C2e6xy = C _ { 1 } e ^ { x } + C _ { 2 } e ^ { - 6 x }
E) None of these
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52
Solve the initial-value problem. y2y24y=0,y(1)=4,y(1)=5y ^ { \prime \prime } - 2 y ^ { \prime } - 24 y = 0 , y ( 1 ) = 4 , y ^ { \prime } ( 1 ) = 5 .

A) y(x)=110e4(x1)+123e6(x1)y ( x ) = \frac { 1 } { 10 } e ^ { 4 ( x - 1 ) } + \frac { 1 } { 23 } e ^ { - 6 ( x - 1 ) }
B) y(x)=110e4x123e6xy ( x ) = \frac { 1 } { 10 } e ^ { 4 x } - \frac { 1 } { 23 } e ^ { - 6 x }
C) y(x)=e6xe4xy ( x ) = e ^ { 6 x } - e ^ { - 4 x }
D) y(x)=2110e6(x1)+1910e4(x1)y ( x ) = \frac { 21 } { 10 } e ^ { 6 ( x - 1 ) } + \frac { 19 } { 10 } e ^ { - 4 ( x - 1 ) }
E) y(x)=110ex123exy ( x ) = \frac { 1 } { 10 } e ^ { x } - \frac { 1 } { 23 } e ^ { - x }
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53
Solve the initial-value problem. Solve the initial-value problem.
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54
Find f by solving the initial value problem. f(x)=8x2+6x+4f ^ { \prime \prime } ( x ) = 8 x ^ { 2 } + 6 x + 4 ; f(1)=7f ( - 1 ) = - 7 , f(1)=3f ^ { \prime } ( - 1 ) = - 3

A) 83\frac { 8 } { 3 } x4x ^ { 4 } +3+ 3 x3x ^ { 3 } + 4 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x - 10
B) 23\frac { 2 } { 3 } x4x ^ { 4 } ++ x3x ^ { 3 } + 2 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x 8- 8
C) 23\frac { 2 } { 3 } x4x ^ { 4 } ++ x3x ^ { 3 } + 2 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x 16- 16
D) 83\frac { 8 } { 3 } x4x ^ { 4 } +3+ 3 x3x ^ { 3 } + 4 x2x ^ { 2 } +23+ \frac { 2 } { 3 } x - 5
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55
Solve the boundary-value problem, if possible. y+16y+64y=0,y(0)=0,y(1)=4y ^ { \prime \prime } + 16 y ^ { \prime } + 64 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 4

A) y(x)=4xe(8x+8)y ( x ) = 4 x e ^ { ( - 8 x + 8 ) }
B) y(x)=4e(8x8)y ( x ) = 4 e ^ { ( - 8 x - 8 ) }
C) y(x)=4xe(8x8)y ( x ) = 4 x e ^ { ( - 8 x - 8 ) }
D) y(x)=8xe(4x+4)y ( x ) = 8 x e ^ { ( - 4 x + 4 ) }
E) No solution
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56
Solve the initial-value problem. y+8y+41y=0,y(0)=1,y(0)=7y ^ { \prime \prime } + 8 y ^ { \prime } + 41 y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 7

A) y=e5x(cos(5x)+115sin(5x))y = e ^ { 5 x } \left( \cos ( 5 x ) + \frac { 11 } { 5 } \sin ( 5 x ) \right)
B) y=e5x(cos(5x)+25sin(5x))y = e ^ { - 5 x } \left( \cos ( 5 x ) + \frac { 2 } { 5 } \sin ( 5 x ) \right)
C) y=e5x(cos(4x)+115sin(4x))y = e ^ { 5 x } \left( \cos ( 4 x ) + \frac { 11 } { 5 } \sin ( 4 x ) \right)
D) y=e4x(cos(5x)+115sin(5x))y = e ^ { - 4 x } \left( \cos ( 5 x ) + \frac { 11 } { 5 } \sin ( 5 x ) \right)
E) y=e4x(cos(5x)25sin(5x))y = e ^ { 4 x } \left( \cos ( 5 x ) - \frac { 2 } { 5 } \sin ( 5 x ) \right)
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57
Solve the initial-value problem. y+81y=0,y(π9)=0,y(π9)=8y ^ { \prime \prime } + 81 y = 0 , y \left( \frac { \pi } { 9 } \right) = 0 , y ^ { \prime } \left( \frac { \pi } { 9 } \right) = 8

A) y(x)=xsin(9x)y ( x ) = - x \sin ( 9 x )
B) y(x)=89xsin(9x)y ( x ) = \frac { 8 } { 9 } x \sin ( 9 x )
C) y(x)=89xsin(9x)y ( x ) = - \frac { 8 } { 9 } x \sin ( 9 x )
D) y(x)=89sin(9x)y ( x ) = - \frac { 8 } { 9 } \sin ( 9 x )
E) y(x)=89sin(9x)y ( x ) = \frac { 8 } { 9 } \sin ( 9 x )
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58
Solve the initial value problem. Solve the initial value problem.
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59
Solve the initial-value problem. y+2y3y=0,y(0)=2,y(0)=1y ^ { \prime \prime } + 2 y ^ { \prime } - 3 y = 0 , y ( 0 ) = 2 , y ^ { \prime } ( 0 ) = 1

A) y=14e2x+74e2xy = \frac { 1 } { 4 } e ^ { 2 x } + \frac { 7 } { 4 } e ^ { - 2 x }
B) y=ex+14e3xy = e ^ { x } + \frac { 1 } { 4 } e ^ { - 3 x }
C) y=e2x+14e2xy = e ^ { 2 x } + \frac { 1 } { 4 } e ^ { - 2 x }
D) y=74ex+14e3xy = \frac { 7 } { 4 } e ^ { x } + \frac { 1 } { 4 } e ^ { - 3 x }
E) None of these
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60
Solve the differential equation. Solve the differential equation.
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61
Solve the boundary-value problem, if possible. Solve the boundary-value problem, if possible.
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62
Solve the initial-value problem. Solve the initial-value problem.
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63
Solve the differential equation. Solve the differential equation.
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