Deck 16: Mathematics Problems: Differential Equations and Linear Algebra

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Question
The solution of y4y+13y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 13 y = 0 is

A) y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { - 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { - 2 x } \sin ( 3 x )
B) y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { - 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { 2 x } \sin ( 3 x )
C) y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { 2 x } \sin ( 3 x )
D) y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 3 x }
E) y=c1cos(3x)+c2sin(3x)y = c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x )
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Question
The solution of y5y+6y=0y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0 is

A) y=c1e2x+c2e3xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x }
B) y=c1e2x+c2xe3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } x e ^ { 3 x }
C) y=c1e2x+c2xe3xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } x e ^ { - 3 x }
D) y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 3 x }
E) y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { - 3 x }
Question
The solution of xy=(x+1)y2x y ^ { \prime } = ( x + 1 ) y ^ { 2 }

A) y=1/(x+lnx+c)y = 1 / ( x + \ln x + c )
B) y=1/(xlnx+c)y = 1 / ( x - \ln x + c )
C) y=c/(x+lnx)y = - c / ( x + \ln x )
D) y=c/(xlnx)y = - c / ( x - \ln x )
E) y=1/(x+lnx+c)y = - 1 / ( x + \ln x + c )
Question
The correct form of the particular solution of y2y=x+exy ^ { \prime \prime } - 2 y ^ { \prime } = x + e ^ { x } is

A) yp=Ax+B+Cexy _ { p } = A x + B + C e ^ { x }
B) yp=(Ax+B+Cex)xy _ { p } = \left( A x + B + C e ^ { x } \right) x
C) yp=Ax2+B+Cexy _ { p } = A x ^ { 2 } + B + C e ^ { x }
D) yp=Ax2+Bx+Cexy _ { p } = A x ^ { 2 } + B x + C e ^ { x }
E) yp=Ax+B+Cxexy _ { p } = A x + B + C x e ^ { x }
Question
The solution of y+y=tanxy ^ { \prime \prime } + y = \tan x is

A) y=c1cosx+c2sinx+cosxlnsecx+tanxy = c _ { 1 } \cos x + c _ { 2 } \sin x + \cos x \ln | \sec x + \tan x |
B) y=c1cosx+c2sinxcosxlnsecx+tanxy = c _ { 1 } \cos x + c _ { 2 } \sin x - \cos x \ln | \sec x + \tan x |
C) y=c1cosx+c2sinx+cosxlnsecxy = c _ { 1 } \cos x + c _ { 2 } \sin x + \cos x \ln | \sec x |
D) y=c1cosx+c2sinxcosxlntanxy = c _ { 1 } \cos x + c _ { 2 } \sin x - \cos x \ln | \tan x |
E) y=c1cosx+c2sinxcosxlnsecxtanxy = c _ { 1 } \cos x + c _ { 2 } \sin x - \cos x \ln | \sec x - \tan x |
Question
The solution of x2ydx+(x3/3+y)dy=0x ^ { 2 } y d x + \left( x ^ { 3 } / 3 + y \right) d y = 0 is

A) x3y/3+y2/2x ^ { 3 } y / 3 + y ^ { 2 } / 2
B) x3y/3y2/2x ^ { 3 } y / 3 - y ^ { 2 } / 2
C) x3y/3+y2/2cx ^ { 3 } y / 3 + y ^ { 2 } / 2 - c
D) x3y/3+y2/2=cx ^ { 3 } y / 3 + y ^ { 2 } / 2 = c
E) x3y/3y2/2=cx ^ { 3 } y / 3 - y ^ { 2 } / 2 = c
Question
The solution of y+3y4y=cosxy ^ { \prime \prime } + 3 y ^ { \prime } - 4 y = \cos x is

A) y=c1ex+c2e4x+(5sinx+3cosx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( 5 \sin x + 3 \cos x ) / 34
B) y=c1ex+c2e4x+(5sinx+3cosx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( - 5 \sin x + 3 \cos x ) / 34
C) y=c1ex+c2e4x+(5cosx3sinx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( - 5 \cos x - 3 \sin x ) / 34
D) y=c1ex+c2e4x+(5cosx+3sinx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( 5 \cos x + 3 \sin x ) / 34
E) y=c1ex+c2e4x+(5cosx+3sinx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( - 5 \cos x + 3 \sin x ) / 34
Question
The solution of yy=xy ^ { \prime } - y = x is

A) y=x1+cexy = x - 1 + c e ^ { x }
B) y=x+1+cexy = - x + 1 + c e ^ { x }
C) y=x1+cexy = - x - 1 + c e ^ { x }
D) y=x1+cexy = - x - 1 + c e ^ { - x }
E) y=x+1+cexy = x + 1 + c e ^ { - x }
Question
The auxiliary equation of y5y+6y=0y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0 is

A) m25m6=0m ^ { 2 } - 5 m - 6 = 0
B) m25m+6=0m ^ { 2 } - 5 m + 6 = 0
C) m25m+6=1m ^ { 2 } - 5 m + 6 = 1
D) m25m+6m ^ { 2 } - 5 m + 6
E) m25m6m ^ { 2 } - 5 m - 6
Question
The solution of x2y+3xy3y=0x ^ { 2 } y ^ { \prime \prime } + 3 x y ^ { \prime } - 3 y = 0 is

A) y=c1x+c2x3y = c _ { 1 } x + c _ { 2 } x ^ { - 3 }
B) y=c1x1+c2x3y = c _ { 1 } x ^ { - 1 } + c _ { 2 } x ^ { 3 }
C) y=c1ex+c2e3xy = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 3 x }
D) y=c1ex+c2e3xy = c _ { 1 } e ^ { - x } + c _ { 2 } e ^ { 3 x }
E) y=c1x+c2x3y = c _ { 1 } x + c _ { 2 } x ^ { 3 }
Question
The solution of y+4y+4y=0y ^ { \prime \prime } + 4 y ^ { \prime } + 4 y = 0 is

A) y=c1e2x+c2xe2xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } x e ^ { - 2 x }
B) y=c1e2x+c2e2xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 2 x }
C) y=c1e2x+c2e2xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 2 x }
D) y=c1e2x+c2xe2xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } x e ^ { 2 x }
E) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
Question
The solution of y2y=x+exy ^ { \prime \prime } - 2 y ^ { \prime } = x + e ^ { x } is

A) y=c1+c2e2xx2/4x/4exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } - x ^ { 2 } / 4 - x / 4 - e ^ { x }
B) y=c1+c2e2xx2/4x/4+exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } - x ^ { 2 } / 4 - x / 4 + e ^ { x }
C) y=c1+c2e2x+x2/4x/4exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } + x ^ { 2 } / 4 - x / 4 - e ^ { x }
D) y=c1+c2e2x+x2/4+x/4exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } + x ^ { 2 } / 4 + x / 4 - e ^ { x }
E) y=c1+c2e2x+x2/4+x/4+exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } + x ^ { 2 } / 4 + x / 4 + e ^ { x }
Question
The correct form of the particular solution of y2y+y=exy ^ { \prime \prime } - 2 y ^ { \prime } + y = e ^ { x } is

A) yp=Aexy _ { p} = A e ^ { x }
B) yp=Axexy _ { p } = A x e ^ { x }
C) yp=Ax2exy _ { p } = A x ^ { 2 } e ^ { x }
D) yp=Ax3exy _ { p } = A x ^ { 3 } e ^ { x }
E) none of the above
Question
A frozen chicken at 0C0 ^ { \circ } \mathrm { C } is taken out of the freezer and placed on a table at 20C20 ^ { \circ } \mathrm { C } . One hour later the temperature of the chicken is 18C18 ^ { \circ } \mathrm { C } . The mathematical model for the temperature T(t)T ( t ) as a function of time tt is (assuming Newton's law of warming)

A) dTdt=kT,T(0)=0,T(1)=18\frac { d T } { d t } = k T , T ( 0 ) = 0 , T ( 1 ) = 18
B) dTdt=k(T20),T(0)=0,T(1)=18\frac { d T } { d t } = k ( T - 20 ) , T ( 0 ) = 0 , T ( 1 ) = 18
C) dTdt=(T20),T(0)=0,T(1)=18\frac { d T } { d t } = ( T - 20 ) , T ( 0 ) = 0 , T ( 1 ) = 18
D) dTdt=T,T(0)=0,T(1)=18\frac { d T } { d t } = T , T ( 0 ) = 0 , T ( 1 ) = 18
E) dTdt=k(T18),T(0)=0,T(1)=18\frac { d T } { d t } = k ( T - 18 ) , T ( 0 ) = 0 , T ( 1 ) = 18
Question
The solution of x2y+xy=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } = 0 is

A) y=c1+c2x1y = c _ { 1 } + c _ { 2 } x ^ { - 1 }
B) y=c1lnx+c2x1y = c _ { 1 } \ln x + c _ { 2 } x ^ { - 1 }
C) y=c1+c2lnxy = c _ { 1 } + c _ { 2 } \ln x
D) y=c1+c2xy = c _ { 1 } + c _ { 2 } x
E) y=c1+c2x2y = c _ { 1 } + c _ { 2 } x ^ { - 2 }
Question
A 2-pound weight is hung on a spring and stretches it 1/2 foot. The mass spring system is then put into motion in a medium offering a damping force numerically equal to the velocity. If the mass is pulled down 4 inches from equilibrium and released, the initial value problem describing the position, x(t)x ( t ) , of the mass at time tt is

A) x16xt+64x=0,x(0)=4,x(0)=0x ^ { \prime \prime } - 16 x ^ { t } + 64 x = 0 , x ( 0 ) = 4 , x ^ { \prime } ( 0 ) = 0
B) x+16xt+64x=0,x(0)=4,x(0)=0x ^ { \prime \prime } + 16 x ^ { t } + 64 x = 0 , x ( 0 ) = 4 , x ^ { \prime } ( 0 ) = 0
C) x16x+64x=0,x(0)=1/3,x(0)=0x ^ { \prime \prime } - 16 x ^ { \prime } + 64 x = 0 , x ( 0 ) = 1 / 3 , x ^ { \prime } ( 0 ) = 0
D) x+16x+64x=0,x(0)=1/3,x(0)=0x ^ { \prime \prime } + 16 x ^ { \prime } + 64 x = 0 , x ( 0 ) = 1 / 3 , x ^ { \prime } ( 0 ) = 0
E) x+64x=16,x(0)=1/3,x(0)=0x ^ { \prime \prime } + 64 x = 16 , x ( 0 ) = 1 / 3 , x ^ { \prime } ( 0 ) = 0
Question
In the previous problem, the solution of the differential equation is

A) T=CektT = C e ^ { k t }
B) T=CektT = C e ^ { - k t }
C) T=20+CektT = 20 + C e ^ { k t }
D) T=20+CektT = 20 + C e ^ { - k t }
E) T=18+CektT = 18 + C e ^ { k t }
Question
In the previous two problems, the solution for the temperature is

A) T(t)=2020e230tT ( t ) = 20 - 20 e ^ { - 230 t }
B) T(t)=2020e230tT ( t ) = 20 - 20 e ^ { 230 t }
C) T(t)=1818e230tT ( t ) = 18 - 18 e ^ { - 230 t }
D) T(t)=1818e230tT ( t ) = 18 - 18 e ^ { 230 t }
E) T(t)=18e230tT ( t ) = 18 e ^ { - 230 t }
Question
In the previous problem, the solution for the position, x(t)x ( t ) , is

A) x=(e8t+8te8t)/3x = \left( e ^ { 8 t } + 8 t e ^ { 8 t } \right) / 3
B) x=(e8t+8te8t)/3x = \left( e ^ { - 8 t } + 8 t e ^ { - 8 t } \right) / 3
C) x=cos(8t)/12+1/4x = \cos ( 8 t ) / 12 + 1 / 4
D) x=(4e8t+32te8t)x = \left( 4 e ^ { - 8 t } + 32 t e ^ { - 8 t } \right)
E) x=(4e8t32te8t)x = \left( 4 e ^ { 8 t } - 32 t e ^ { 8 t } \right)
Question
The solution of yy=(y)2y y ^ { \prime \prime } = \left( y ^ { \prime } \right) ^ { 2 } is

A) y=c1eC2xy = c _ { 1 } e ^ { C _ { 2 } x }
B) yln(c1)+ylnyy=x+c2y \ln \left( c _ { 1 } \right) + y \ln y - y = - x + c _ { 2 }
C) y=c1x+c2y = c _ { 1 } x + c _ { 2 }
D) ylnyy=c1x+c2y \ln y - y = c _ { 1 } x + c _ { 2 }
E) y=c1ln(c2x)y = c _ { 1 } \ln \left( c _ { 2 } x \right)
Question
Using Laplace transform methods, the solution of y+y=2sint,y(0)=1y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1 is (Hint: the previous problem might be useful.)

A) y=2et+sint+costy = 2 e ^ { - t } + \sin t + \cos t
B) y=et+etsintcosty = e ^ { t } + e ^ { - t } - \sin t - \cos t
C) y=2etsintcosty = 2 e ^ { - t } - \sin t - \cos t
D) y=2et+sintcosty = 2 e ^ { - t } + \sin t - \cos t
E) y=et+et+sintcosty = e ^ { t } + e ^ { - t } + \sin t - \cos t
Question
Let A=(2003)A = \left( \begin{array} { c c } 2 & 0 \\0 & - 3\end{array} \right) . Then eAt=e ^ { A t } =

A) I+AtI + A t
B) I+At+A2t2/2I + A t + A ^ { 2 } t ^ { 2 } / 2
C) At+A2t2/2A t + A ^ { 2 } t ^ { 2 } / 2
D) (e2t00e3t)\left( \begin{array} { c c } e ^ { 2 t } & 0 \\0 & e ^ { - 3 t }\end{array} \right)
E) (e2t1100e3t1)\left( \begin{array} { c c } e ^ { 2 t - 1 } - 1 & 0 \\0 & e ^ { - 3 t } - 1\end{array} \right)
Question
The solution of the previous problem is

A) y=w0[4(t5)3u(t5)+15t22t3]/(24EI)y = w _ { 0 } \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) + 15 t ^ { 2 } - 2 t ^ { 3 } \right] / ( 24 E I )
B) y=w0[4(t5)3u(t5)+15t22t3]/(12EI)y = w _ { 0 } \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) + 15 t ^ { 2 } - 2 t ^ { 3 } \right] / ( 12 E I )
C) y=w0EI[4(t5)3u(t5)+15t22t3]/24y = w _ { 0 } E I \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) + 15 t ^ { 2 } - 2 t ^ { 3 } \right] / 24
D) y=w0EI[4(t5)3u(t5)15t22t3]/12y = w _ { 0 } E I \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) - 15 t ^ { 2 } - 2 t ^ { 3 } \right] / 12
E) y=w0[4(t5)3u(t5)15t22t3]/(24EI)y = w _ { 0 } \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) - 15 t ^ { 2 } - 2 t ^ { 3 } \right] / ( 24 E I )
Question
Using Laplace transform methods, the solution of y+2y+y=et,y(0)=1,y(0)=0y ^ { \prime \prime } + 2 y ^ { \prime } + y = e ^ { - t } , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is

A) y=et+tet+t2et/2y = e ^ { t } + t e ^ { t } + t ^ { 2 } e ^ { t } / 2
B) y=et+tet+t2et/2y = e ^ { - t } + t e ^ { - t } + t ^ { 2 } e ^ { - t } / 2
C) y=ettet+t2et/2y = e ^ { t } - t e ^ { t } + t ^ { 2 } e ^ { t } / 2
D) y=ettett2et/2y = e ^ { - t } - t e ^ { - t } - t ^ { 2 } e ^ { - t } / 2
E) y=et+tett2et/2y = e ^ { - t } + t e ^ { - t } - t ^ { 2 } e ^ { - t } / 2
Question
In the previous problem, the exact solution of the initial value problem is

A) y=(e2x1)/(e2x+1)y = \left( e ^ { 2 x } - 1 \right) / \left( e ^ { 2 x } + 1 \right)
B) y=(e2x+1)/(e2x1)y = \left( e ^ { 2 x } + 1 \right) / \left( e ^ { 2 x } - 1 \right)
C) y=(e2x1)/(e2x+1)y = \left( e ^ { - 2 x } - 1 \right) / \left( e ^ { - 2 x } + 1 \right)
D) y=(e2x+1)/(e2x1)y = - \left( e ^ { - 2 x } + 1 \right) / \left( e ^ { - 2 x } - 1 \right)
E) y=(e2x1)/(e2x+1)y = - \left( e ^ { 2 x } - 1 \right) / \left( e ^ { 2 x } + 1 \right)
Question
The eigenvalues of the matrix A=(1214)A = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) are

A) (5±17)/2( 5 \pm \sqrt { 17 } ) / 2
B) (5±17)/2( - 5 \pm \sqrt { 17 } ) / 2
C) 1, 2
D) 2, 3
E) 2, 2
Question
The solution of X=(300031011)XX ^ { \prime } = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X } is

A) X=c1(010)e3t+c2(011)e2t+c3[(011)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
B) X=c1(011)e3t+c2(010)e2t+c3[(010)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
C) X=c1(100)e3t+c2(011)e2t+c3[(011)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
D) X=c1(100)e3t+c2(011)e2t+c3[(011)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l } 0 \\1 \\1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 0 \\1 \\1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
E) X=c1(100)e3t+c2(110)e2t+c3[(110)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
Question
The solution of X=(1214)X\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) \mathbf { X } is

A) X=c1(21)e2t+c2(11)e3t\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 2 \\1\end{array} \right) e ^ { - 2 t } + c _ { 2 } \left( \begin{array} { l } 1 \\1\end{array} \right) e ^ { - 3 t }
B) X=c1(21)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { l } 2 \\1\end{array} \right) e ^ { - 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { - 3 t }
C) X=c1(21)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { c } - 2 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
D) X=c1(12)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { c } 1 \\- 2\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
E) X=c1(21)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { c } 2 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
Question
A particular solution of X=(1211)X+(2t)\mathbf { X } ^ { \prime } = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) \mathbf { X } + \left( \begin{array} { l } 2 \\t\end{array} \right) is

A) Xp=(2t+2t+3)\mathrm { X } _ { \boldsymbol { p } } = \left( \begin{array} { c } 2 t + 2 \\- t + 3\end{array} \right)
B) Xp=(2t+2t+3)\mathrm { X } _ { p } = \left( \begin{array} { c } - 2 t + 2 \\- t + 3\end{array} \right)
C) Xp=(2t+2t3)X _ { p } = \left( \begin{array} { c } - 2 t + 2 \\- t - 3\end{array} \right)
D) Xy=(2t+2t+3)\mathrm { X } _ { y } = \left( \begin{array} { c } - 2 t + 2 \\t + 3\end{array} \right)
E) Xy=(2t2t3)X _ { y } = \left( \begin{array} { c } - 2 t - 2 \\- t - 3\end{array} \right)
Question
Using the improved Euler method with a step size of h=0.1h = 0.1 , the solution of y=1y2,y(0)=0y ^ { \prime } = 1 - y ^ { 2 } , y ( 0 ) = 0 at x=0.1x = 0.1 is

A) y1=0.095y _ { 1 } = 0.095
B) y1=0.995y _ { 1 } = 0.995
C) y1=0.95y _ { 1 } = 0.95
D) y1=0.00995y _ { 1 } = 0.00995
E) y1=0.0995y _ { 1 } = 0.0995
Question
Using the convolution theorem, we find that L1{1/((s+1)(s2+1))}=\mathcal { L } ^ { - 1 } \left\{ 1 / \left( ( s + 1 ) \left( s ^ { 2 } + 1 \right) \right) \right\} =

A) (et+sintcost)/2\left( e ^ { - t } + \sin t - \cos t \right) / 2
B) (et+sintcost)/2\left( e ^ { t } + \sin t - \cos t \right) / 2
C) (et+sint+cost)/2\left( e ^ { - t } + \sin t + \cos t \right) / 2
D) (etsintcost)/2\left( e ^ { t } - \sin t - \cos t \right) / 2
E) (etsintcost)/2\left( e ^ { - t } - \sin t - \cos t \right) / 2
Question
The solution of X=(1211)X\mathbf { X } ^ { \prime } = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) \mathbf { X } is

A) X=c1[(21)cost(01)sint]+c2[(21)sint+(01)cost]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \cos t - \left( \begin{array} { c } 0 \\- 1\end{array} \right) \sin t \right] + c _ { 2 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \sin t + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos t \right]
B) X=c1(10)e3t+c2(01)e3tX = c _ { 1 } \left( \begin{array} { l } 1 \\0\end{array} \right) e ^ { \sqrt { 3 } t } + c _ { 2 } \left( \begin{array} { l } 0 \\1\end{array} \right) e ^ { - \sqrt { 3 } t }
C) X=c1(10)et+c2(01)et\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0\end{array} \right) e ^ { t } + c _ { 2 } \left( \begin{array} { l } 0 \\1\end{array} \right) e ^ { - t }
D) X=c1[(21)cos(3t)(01)sin(3t)]+c2[(21)sin(3t)+(01)cos(3t)]\begin{array} { l } \mathrm { X } = c _ { 1 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \cos ( \sqrt { 3 } t ) - \left( \begin{array} { c } 0 \\- 1\end{array} \right) \sin ( \sqrt { 3 } t ) \right] + \\c _ { 2 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \sin ( \sqrt { 3 } t ) + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos ( \sqrt { 3 } t ) \right]\end{array}
E) X=c1[(21)cos(2t)(01)sin(2t)]+c2[(21)sin(2t)+(01)cos(2t)]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \cos ( 2 t ) - \left( \begin{array} { c } 0 \\- 1\end{array} \right) \sin ( 2 t ) \right] + c _ { 2 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \sin ( 2 t ) + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos ( 2 t ) \right]
Question
In the previous two problems, the error in the improved Euler method at x=0.1x = 0.1 is

A) 0.00467
B) 0.000168
C) 0.870
D) 0.895
E) 0.0897
Question
Using Laplace transform methods, the solution of y+y=δ(tπ),y(0)=1,y(0)=0y ^ { \prime \prime } + y = \delta ( t - \pi ) , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is

A) y=sint+sin(tπ)u(tπ)y = \sin t + \sin ( t - \pi ) \boldsymbol { u } ( t - \pi )
B) y=sintcos(tπ)u(tπ)y = \sin t - \cos ( t - \pi ) u ( t - \pi )
C) y=cost+sin(tπ)u(tπ)y = \cos t + \sin ( t - \pi ) u ( t - \pi )
D) y=cost+cos(tπ)u(tπ)y = \cos t + \cos ( t - \pi ) u ( t - \pi )
E) y=costsin(tπ)u(tπ)y = \cos t - \sin ( t - \pi ) u ( t - \pi )
Question
A uniform beam of length 10 has a concentrated load w0w _ { 0 } at x=5x = 5 . It is embedded at both ends. The boundary value problem for the deflections, y(x)y ( x ) , for this system is

A) y=ELw0δ(x5),y(0)=0,y(0)=0,y(10)=0,y(10)=0y ^ { \prime \prime \prime \prime } = E L w _ { 0 } \delta ( x - 5 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
B) y=Elw0δ(x10),y(0)=0,y(0)=0,y(10)=0,y(10)=0y ^ { \prime \prime } = E l w _ { 0 } \delta ( x - 10 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
C) Ely=w0δ(x5),y(0)=0,y(0)=0,y(10)=0,y(10)=0E l y ^ { \prime \prime } = w _ { 0 } \delta ( x - 5 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
D) ELy=w0δ(x5),y(0)=0,y(0)=0,y(10)=0,y(10)=0E L y ^ { \prime \prime \prime } = w _ { 0 } \delta ( x - 5 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
E) ELy=w0δ(x10),y(0)=0,y(0)=0,y(10)=0,y(10)=0E L y ^ { \prime \prime \prime \prime } = w _ { 0 } \delta ( x - 10 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
Question
The eigenvalues of the matrix A=(300031011)A = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) are

A) 1, 2, 3
B) 2, 2, 3
C) 1, 2, 2
D) 2,2,3- 2 , - 2,3
E) 1,2,3- 1 , - 2,3
Question
Using power series methods, the solution of 2xy+y+2y=02 x y ^ { \prime \prime } + y ^ { \prime } + 2 y = 0 is

A) y=c0n1(2)nxn/(n!(13(2n1)))+c1x1/2n1(2)nxn/(n!(35(2n+1)))\begin{array} { l } y = c _ { 0 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) + \\c _ { 1 } x ^ { 1 / 2 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) )\end{array}
B) y=c0n1(2)nxn/(n!(13(2n1)))+c1x1/2[1+n1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = c _ { 0 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) + \\c _ { 1 } x ^ { 1 / 2 } \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
C) y=c0[1+n=1(2)nxn/(n!(13(2n1)))]+c1[1+n=1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = c _ { 0 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\c _ { 1 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
D) y=c0[1+n=1(2)nxn/(n!(13(2n1)))]+c1x1/2[1+n=1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = c _ { 0 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\c _ { 1 } x ^ { 1 / 2 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
E) y=[1+n1(2)nxn/(n!(13(2n1)))]+x1/2[1+n1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\x ^ { 1 / 2 } \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
Question
Using power series methods, the solution of xyxy+y=0x y ^ { \prime \prime } - x y ^ { \prime } + y = 0 is

A) y=c0x+c1[xlnx1+n=2xn/n!]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n = 2 } ^ { \infty } x ^ { n } / n ! \right]
B) y=c0x+c1[xlnx1+n1xn/(n!(n+1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n - 1 } ^ { \infty } x ^ { n } / ( n ! ( n + 1 ) ) \right]
C) y=c0x+c1[xlnx+n2xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x + \sum _ { n - 2 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
D) y=c0x+c1[xlnx+n1xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x + \sum _ { n - 1 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
E) y=c0x+c1[xlnx1+n2xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n - 2 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
Question
The solution of the eigenvalue problem y+λy=0,y(0)=0,y(1)=0y ^ { \prime \prime } + \lambda y = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 1 ) = 0 is

A) λ=n2π2/4,y=cos(nπx/2),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } / 4 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots
B) λ=nπ/2,y=cos(nπx/2),n=1,2,3,\lambda = n \pi / 2 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots
C) λ=n2π2/4,y=sin(nπx/2),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } / 4 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
D) λ=(2n1)π/2,y=cos((2n1)πx/2),n=1,2,3,\lambda = ( 2 n - 1 ) \pi / 2 , y = \cos ( ( 2 n - 1 ) \pi x / 2 ) , n = 1,2,3 , \ldots
E) λ=(2n1)2π2/4,y=cos((2n1)πx/2),n=1,2,3,\lambda = ( 2 n - 1 ) ^ { 2 } \pi ^ { 2 } / 4 , y = \cos ( ( 2 n - 1 ) \pi x / 2 ) , n = 1,2,3 , \ldots
Question
The eigenvalues of the matrix A=(1211)A = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) are

A) ±3\pm \sqrt { 3 }
B) ±3i\pm \sqrt { 3 } i
C) ±1\pm 1
D) ±2i\pm 2 i
E) ±\pm
Question
In the previous problem, the error in the classical Runge-Kutta method at x=0.1x = 0.1 is (Hint: see the previous five problems.)

A) 0.0008
B) 0.00008
C) 0.00000008
D) 0.000008
E) 0.0000008
Question
The eigenvalues of the matrix A=(300031011)A = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) are

A) (100)\left( \begin{array} { l } 1 \\0 \\0\end{array} \right)
B) (011)\left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right)
C) (110)\left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right)
D) (011)\left( \begin{array} { l } 0 \\1 \\1\end{array} \right)
E) (010)\left( \begin{array} { l } 0 \\1 \\0\end{array} \right)
Question
The differential equation y=x2y2y ^ { \prime } = x ^ { 2 } y ^ { 2 } is Select all that apply.

A) linear
B) separable
C) exact
D) non-linear
E) Bernoulli
Question
Consider the boundary-value problem y4y+3y=x,y(0)=1,y(1)=2y ^ { \prime \prime } - 4 y ^ { \prime } + 3 y = x , y ( 0 ) = 1 , y ( 1 ) = 2 . Replace the derivatives with central differences with a step size of h=1/4h = 1 / 4 . The resulting equations are

A) 12yi+129yi24yi1=xi12 y _ { i + 1 } - 29 y _ { i } - 24 y _ { i - 1 } = x _ { i }
B) 12yi+1+17yi+12yi1=xi12 y _ { i + 1 } + 17 y _ { i } + 12 y _ { i - 1 } = x _ { i }
C) 8yi+1+17yi+12yi1=xi8 y _ { i + 1 } + 17 y _ { i } + 12 y _ { i - 1 } = x _ { i }
D) 8yi+129yi+24yi1=xi8 y _ { i + 1 } - 29 y _ { i } + 24 y _ { i - 1 } = x _ { i }
E) 8yi+1+29yi24yi1=xi8 y _ { i + 1 } + 29 y _ { i } - 24 y _ { i - 1 } = x _ { i }
Question
Using the classical Runge-Kutta method of order 4 with a step size of h=0.1h = 0.1 , the solution of y=1y2,y(0)=0y ^ { \prime } = 1 - y ^ { 2 } , y ( 0 ) = 0 at x=0.1x = 0.1 is

A) 0.099588
B) 0.099668
C) 0.099688
D) 0.099768
E) 0.099788
Question
The differential equation xdyydx=0x d y - y d x = 0 is Select all that apply.

A) linear
B) separable
C) exact
D) non-linear
E) Bernoulli
Question
The differential equation y+y=x2y ^ { \prime } + y = x ^ { 2 } is Select all that apply.

A) linear
B) separable
C) exact
D) non-linear
E) Bernoulli
Question
The solution of the system in the previous problem is

A) y1=1.228,y2=1.482,y3=1.753y _ { 1 } = 1.228 , y _ { 2 } = 1.482 , y _ { 3 } = 1.753
B) y1=1.228,y2=1.646,y3=1.753y _ { 1 } = 1.228 , y _ { 2 } = 1.646 , y _ { 3 } = 1.753
C) y1=1.126,y2=1.646,y3=2.903y _ { 1 } = 1.126 , y _ { 2 } = 1.646 , y _ { 3 } = 2.903
D) y1=1.126,y2=1.786,y3=2.903y _ { 1 } = 1.126 , y _ { 2 } = 1.786 , y _ { 3 } = 2.903
E) y1=1.016,y2=1.786,y3=2.903y _ { 1 } = 1.016 , y _ { 2 } = 1.786 , y _ { 3 } = 2.903
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Deck 16: Mathematics Problems: Differential Equations and Linear Algebra
1
The solution of y4y+13y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 13 y = 0 is

A) y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { - 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { - 2 x } \sin ( 3 x )
B) y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { - 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { 2 x } \sin ( 3 x )
C) y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { 2 x } \sin ( 3 x )
D) y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 3 x }
E) y=c1cos(3x)+c2sin(3x)y = c _ { 1 } \cos ( 3 x ) + c _ { 2 } \sin ( 3 x )
y=c1e2xcos(3x)+c2e2xsin(3x)y = c _ { 1 } e ^ { 2 x } \cos ( 3 x ) + c _ { 2 } e ^ { 2 x } \sin ( 3 x )
2
The solution of y5y+6y=0y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0 is

A) y=c1e2x+c2e3xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 3 x }
B) y=c1e2x+c2xe3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } x e ^ { 3 x }
C) y=c1e2x+c2xe3xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } x e ^ { - 3 x }
D) y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 3 x }
E) y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { - 3 x }
y=c1e2x+c2e3xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 3 x }
3
The solution of xy=(x+1)y2x y ^ { \prime } = ( x + 1 ) y ^ { 2 }

A) y=1/(x+lnx+c)y = 1 / ( x + \ln x + c )
B) y=1/(xlnx+c)y = 1 / ( x - \ln x + c )
C) y=c/(x+lnx)y = - c / ( x + \ln x )
D) y=c/(xlnx)y = - c / ( x - \ln x )
E) y=1/(x+lnx+c)y = - 1 / ( x + \ln x + c )
y=1/(x+lnx+c)y = - 1 / ( x + \ln x + c )
4
The correct form of the particular solution of y2y=x+exy ^ { \prime \prime } - 2 y ^ { \prime } = x + e ^ { x } is

A) yp=Ax+B+Cexy _ { p } = A x + B + C e ^ { x }
B) yp=(Ax+B+Cex)xy _ { p } = \left( A x + B + C e ^ { x } \right) x
C) yp=Ax2+B+Cexy _ { p } = A x ^ { 2 } + B + C e ^ { x }
D) yp=Ax2+Bx+Cexy _ { p } = A x ^ { 2 } + B x + C e ^ { x }
E) yp=Ax+B+Cxexy _ { p } = A x + B + C x e ^ { x }
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5
The solution of y+y=tanxy ^ { \prime \prime } + y = \tan x is

A) y=c1cosx+c2sinx+cosxlnsecx+tanxy = c _ { 1 } \cos x + c _ { 2 } \sin x + \cos x \ln | \sec x + \tan x |
B) y=c1cosx+c2sinxcosxlnsecx+tanxy = c _ { 1 } \cos x + c _ { 2 } \sin x - \cos x \ln | \sec x + \tan x |
C) y=c1cosx+c2sinx+cosxlnsecxy = c _ { 1 } \cos x + c _ { 2 } \sin x + \cos x \ln | \sec x |
D) y=c1cosx+c2sinxcosxlntanxy = c _ { 1 } \cos x + c _ { 2 } \sin x - \cos x \ln | \tan x |
E) y=c1cosx+c2sinxcosxlnsecxtanxy = c _ { 1 } \cos x + c _ { 2 } \sin x - \cos x \ln | \sec x - \tan x |
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6
The solution of x2ydx+(x3/3+y)dy=0x ^ { 2 } y d x + \left( x ^ { 3 } / 3 + y \right) d y = 0 is

A) x3y/3+y2/2x ^ { 3 } y / 3 + y ^ { 2 } / 2
B) x3y/3y2/2x ^ { 3 } y / 3 - y ^ { 2 } / 2
C) x3y/3+y2/2cx ^ { 3 } y / 3 + y ^ { 2 } / 2 - c
D) x3y/3+y2/2=cx ^ { 3 } y / 3 + y ^ { 2 } / 2 = c
E) x3y/3y2/2=cx ^ { 3 } y / 3 - y ^ { 2 } / 2 = c
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7
The solution of y+3y4y=cosxy ^ { \prime \prime } + 3 y ^ { \prime } - 4 y = \cos x is

A) y=c1ex+c2e4x+(5sinx+3cosx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( 5 \sin x + 3 \cos x ) / 34
B) y=c1ex+c2e4x+(5sinx+3cosx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( - 5 \sin x + 3 \cos x ) / 34
C) y=c1ex+c2e4x+(5cosx3sinx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( - 5 \cos x - 3 \sin x ) / 34
D) y=c1ex+c2e4x+(5cosx+3sinx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( 5 \cos x + 3 \sin x ) / 34
E) y=c1ex+c2e4x+(5cosx+3sinx)/34y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 4 x } + ( - 5 \cos x + 3 \sin x ) / 34
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8
The solution of yy=xy ^ { \prime } - y = x is

A) y=x1+cexy = x - 1 + c e ^ { x }
B) y=x+1+cexy = - x + 1 + c e ^ { x }
C) y=x1+cexy = - x - 1 + c e ^ { x }
D) y=x1+cexy = - x - 1 + c e ^ { - x }
E) y=x+1+cexy = x + 1 + c e ^ { - x }
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9
The auxiliary equation of y5y+6y=0y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0 is

A) m25m6=0m ^ { 2 } - 5 m - 6 = 0
B) m25m+6=0m ^ { 2 } - 5 m + 6 = 0
C) m25m+6=1m ^ { 2 } - 5 m + 6 = 1
D) m25m+6m ^ { 2 } - 5 m + 6
E) m25m6m ^ { 2 } - 5 m - 6
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10
The solution of x2y+3xy3y=0x ^ { 2 } y ^ { \prime \prime } + 3 x y ^ { \prime } - 3 y = 0 is

A) y=c1x+c2x3y = c _ { 1 } x + c _ { 2 } x ^ { - 3 }
B) y=c1x1+c2x3y = c _ { 1 } x ^ { - 1 } + c _ { 2 } x ^ { 3 }
C) y=c1ex+c2e3xy = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 3 x }
D) y=c1ex+c2e3xy = c _ { 1 } e ^ { - x } + c _ { 2 } e ^ { 3 x }
E) y=c1x+c2x3y = c _ { 1 } x + c _ { 2 } x ^ { 3 }
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11
The solution of y+4y+4y=0y ^ { \prime \prime } + 4 y ^ { \prime } + 4 y = 0 is

A) y=c1e2x+c2xe2xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } x e ^ { - 2 x }
B) y=c1e2x+c2e2xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 2 x }
C) y=c1e2x+c2e2xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 2 x }
D) y=c1e2x+c2xe2xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } x e ^ { 2 x }
E) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
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12
The solution of y2y=x+exy ^ { \prime \prime } - 2 y ^ { \prime } = x + e ^ { x } is

A) y=c1+c2e2xx2/4x/4exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } - x ^ { 2 } / 4 - x / 4 - e ^ { x }
B) y=c1+c2e2xx2/4x/4+exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } - x ^ { 2 } / 4 - x / 4 + e ^ { x }
C) y=c1+c2e2x+x2/4x/4exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } + x ^ { 2 } / 4 - x / 4 - e ^ { x }
D) y=c1+c2e2x+x2/4+x/4exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } + x ^ { 2 } / 4 + x / 4 - e ^ { x }
E) y=c1+c2e2x+x2/4+x/4+exy = c _ { 1 } + c _ { 2 } e ^ { 2 x } + x ^ { 2 } / 4 + x / 4 + e ^ { x }
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13
The correct form of the particular solution of y2y+y=exy ^ { \prime \prime } - 2 y ^ { \prime } + y = e ^ { x } is

A) yp=Aexy _ { p} = A e ^ { x }
B) yp=Axexy _ { p } = A x e ^ { x }
C) yp=Ax2exy _ { p } = A x ^ { 2 } e ^ { x }
D) yp=Ax3exy _ { p } = A x ^ { 3 } e ^ { x }
E) none of the above
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14
A frozen chicken at 0C0 ^ { \circ } \mathrm { C } is taken out of the freezer and placed on a table at 20C20 ^ { \circ } \mathrm { C } . One hour later the temperature of the chicken is 18C18 ^ { \circ } \mathrm { C } . The mathematical model for the temperature T(t)T ( t ) as a function of time tt is (assuming Newton's law of warming)

A) dTdt=kT,T(0)=0,T(1)=18\frac { d T } { d t } = k T , T ( 0 ) = 0 , T ( 1 ) = 18
B) dTdt=k(T20),T(0)=0,T(1)=18\frac { d T } { d t } = k ( T - 20 ) , T ( 0 ) = 0 , T ( 1 ) = 18
C) dTdt=(T20),T(0)=0,T(1)=18\frac { d T } { d t } = ( T - 20 ) , T ( 0 ) = 0 , T ( 1 ) = 18
D) dTdt=T,T(0)=0,T(1)=18\frac { d T } { d t } = T , T ( 0 ) = 0 , T ( 1 ) = 18
E) dTdt=k(T18),T(0)=0,T(1)=18\frac { d T } { d t } = k ( T - 18 ) , T ( 0 ) = 0 , T ( 1 ) = 18
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15
The solution of x2y+xy=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } = 0 is

A) y=c1+c2x1y = c _ { 1 } + c _ { 2 } x ^ { - 1 }
B) y=c1lnx+c2x1y = c _ { 1 } \ln x + c _ { 2 } x ^ { - 1 }
C) y=c1+c2lnxy = c _ { 1 } + c _ { 2 } \ln x
D) y=c1+c2xy = c _ { 1 } + c _ { 2 } x
E) y=c1+c2x2y = c _ { 1 } + c _ { 2 } x ^ { - 2 }
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16
A 2-pound weight is hung on a spring and stretches it 1/2 foot. The mass spring system is then put into motion in a medium offering a damping force numerically equal to the velocity. If the mass is pulled down 4 inches from equilibrium and released, the initial value problem describing the position, x(t)x ( t ) , of the mass at time tt is

A) x16xt+64x=0,x(0)=4,x(0)=0x ^ { \prime \prime } - 16 x ^ { t } + 64 x = 0 , x ( 0 ) = 4 , x ^ { \prime } ( 0 ) = 0
B) x+16xt+64x=0,x(0)=4,x(0)=0x ^ { \prime \prime } + 16 x ^ { t } + 64 x = 0 , x ( 0 ) = 4 , x ^ { \prime } ( 0 ) = 0
C) x16x+64x=0,x(0)=1/3,x(0)=0x ^ { \prime \prime } - 16 x ^ { \prime } + 64 x = 0 , x ( 0 ) = 1 / 3 , x ^ { \prime } ( 0 ) = 0
D) x+16x+64x=0,x(0)=1/3,x(0)=0x ^ { \prime \prime } + 16 x ^ { \prime } + 64 x = 0 , x ( 0 ) = 1 / 3 , x ^ { \prime } ( 0 ) = 0
E) x+64x=16,x(0)=1/3,x(0)=0x ^ { \prime \prime } + 64 x = 16 , x ( 0 ) = 1 / 3 , x ^ { \prime } ( 0 ) = 0
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17
In the previous problem, the solution of the differential equation is

A) T=CektT = C e ^ { k t }
B) T=CektT = C e ^ { - k t }
C) T=20+CektT = 20 + C e ^ { k t }
D) T=20+CektT = 20 + C e ^ { - k t }
E) T=18+CektT = 18 + C e ^ { k t }
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18
In the previous two problems, the solution for the temperature is

A) T(t)=2020e230tT ( t ) = 20 - 20 e ^ { - 230 t }
B) T(t)=2020e230tT ( t ) = 20 - 20 e ^ { 230 t }
C) T(t)=1818e230tT ( t ) = 18 - 18 e ^ { - 230 t }
D) T(t)=1818e230tT ( t ) = 18 - 18 e ^ { 230 t }
E) T(t)=18e230tT ( t ) = 18 e ^ { - 230 t }
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19
In the previous problem, the solution for the position, x(t)x ( t ) , is

A) x=(e8t+8te8t)/3x = \left( e ^ { 8 t } + 8 t e ^ { 8 t } \right) / 3
B) x=(e8t+8te8t)/3x = \left( e ^ { - 8 t } + 8 t e ^ { - 8 t } \right) / 3
C) x=cos(8t)/12+1/4x = \cos ( 8 t ) / 12 + 1 / 4
D) x=(4e8t+32te8t)x = \left( 4 e ^ { - 8 t } + 32 t e ^ { - 8 t } \right)
E) x=(4e8t32te8t)x = \left( 4 e ^ { 8 t } - 32 t e ^ { 8 t } \right)
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20
The solution of yy=(y)2y y ^ { \prime \prime } = \left( y ^ { \prime } \right) ^ { 2 } is

A) y=c1eC2xy = c _ { 1 } e ^ { C _ { 2 } x }
B) yln(c1)+ylnyy=x+c2y \ln \left( c _ { 1 } \right) + y \ln y - y = - x + c _ { 2 }
C) y=c1x+c2y = c _ { 1 } x + c _ { 2 }
D) ylnyy=c1x+c2y \ln y - y = c _ { 1 } x + c _ { 2 }
E) y=c1ln(c2x)y = c _ { 1 } \ln \left( c _ { 2 } x \right)
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21
Using Laplace transform methods, the solution of y+y=2sint,y(0)=1y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1 is (Hint: the previous problem might be useful.)

A) y=2et+sint+costy = 2 e ^ { - t } + \sin t + \cos t
B) y=et+etsintcosty = e ^ { t } + e ^ { - t } - \sin t - \cos t
C) y=2etsintcosty = 2 e ^ { - t } - \sin t - \cos t
D) y=2et+sintcosty = 2 e ^ { - t } + \sin t - \cos t
E) y=et+et+sintcosty = e ^ { t } + e ^ { - t } + \sin t - \cos t
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22
Let A=(2003)A = \left( \begin{array} { c c } 2 & 0 \\0 & - 3\end{array} \right) . Then eAt=e ^ { A t } =

A) I+AtI + A t
B) I+At+A2t2/2I + A t + A ^ { 2 } t ^ { 2 } / 2
C) At+A2t2/2A t + A ^ { 2 } t ^ { 2 } / 2
D) (e2t00e3t)\left( \begin{array} { c c } e ^ { 2 t } & 0 \\0 & e ^ { - 3 t }\end{array} \right)
E) (e2t1100e3t1)\left( \begin{array} { c c } e ^ { 2 t - 1 } - 1 & 0 \\0 & e ^ { - 3 t } - 1\end{array} \right)
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23
The solution of the previous problem is

A) y=w0[4(t5)3u(t5)+15t22t3]/(24EI)y = w _ { 0 } \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) + 15 t ^ { 2 } - 2 t ^ { 3 } \right] / ( 24 E I )
B) y=w0[4(t5)3u(t5)+15t22t3]/(12EI)y = w _ { 0 } \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) + 15 t ^ { 2 } - 2 t ^ { 3 } \right] / ( 12 E I )
C) y=w0EI[4(t5)3u(t5)+15t22t3]/24y = w _ { 0 } E I \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) + 15 t ^ { 2 } - 2 t ^ { 3 } \right] / 24
D) y=w0EI[4(t5)3u(t5)15t22t3]/12y = w _ { 0 } E I \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) - 15 t ^ { 2 } - 2 t ^ { 3 } \right] / 12
E) y=w0[4(t5)3u(t5)15t22t3]/(24EI)y = w _ { 0 } \left[ 4 ( t - 5 ) ^ { 3 } u ( t - 5 ) - 15 t ^ { 2 } - 2 t ^ { 3 } \right] / ( 24 E I )
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24
Using Laplace transform methods, the solution of y+2y+y=et,y(0)=1,y(0)=0y ^ { \prime \prime } + 2 y ^ { \prime } + y = e ^ { - t } , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is

A) y=et+tet+t2et/2y = e ^ { t } + t e ^ { t } + t ^ { 2 } e ^ { t } / 2
B) y=et+tet+t2et/2y = e ^ { - t } + t e ^ { - t } + t ^ { 2 } e ^ { - t } / 2
C) y=ettet+t2et/2y = e ^ { t } - t e ^ { t } + t ^ { 2 } e ^ { t } / 2
D) y=ettett2et/2y = e ^ { - t } - t e ^ { - t } - t ^ { 2 } e ^ { - t } / 2
E) y=et+tett2et/2y = e ^ { - t } + t e ^ { - t } - t ^ { 2 } e ^ { - t } / 2
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25
In the previous problem, the exact solution of the initial value problem is

A) y=(e2x1)/(e2x+1)y = \left( e ^ { 2 x } - 1 \right) / \left( e ^ { 2 x } + 1 \right)
B) y=(e2x+1)/(e2x1)y = \left( e ^ { 2 x } + 1 \right) / \left( e ^ { 2 x } - 1 \right)
C) y=(e2x1)/(e2x+1)y = \left( e ^ { - 2 x } - 1 \right) / \left( e ^ { - 2 x } + 1 \right)
D) y=(e2x+1)/(e2x1)y = - \left( e ^ { - 2 x } + 1 \right) / \left( e ^ { - 2 x } - 1 \right)
E) y=(e2x1)/(e2x+1)y = - \left( e ^ { 2 x } - 1 \right) / \left( e ^ { 2 x } + 1 \right)
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26
The eigenvalues of the matrix A=(1214)A = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) are

A) (5±17)/2( 5 \pm \sqrt { 17 } ) / 2
B) (5±17)/2( - 5 \pm \sqrt { 17 } ) / 2
C) 1, 2
D) 2, 3
E) 2, 2
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27
The solution of X=(300031011)XX ^ { \prime } = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X } is

A) X=c1(010)e3t+c2(011)e2t+c3[(011)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
B) X=c1(011)e3t+c2(010)e2t+c3[(010)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
C) X=c1(100)e3t+c2(011)e2t+c3[(011)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
D) X=c1(100)e3t+c2(011)e2t+c3[(011)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { l } 0 \\1 \\1\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { l } 0 \\1 \\1\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
E) X=c1(100)e3t+c2(110)e2t+c3[(110)te2t+(010)e2t]\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 3 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right) e ^ { 2 t } + c _ { 3 } \left[ \left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right) t e ^ { 2 t } + \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 2 t } \right]
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28
The solution of X=(1214)X\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) \mathbf { X } is

A) X=c1(21)e2t+c2(11)e3t\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 2 \\1\end{array} \right) e ^ { - 2 t } + c _ { 2 } \left( \begin{array} { l } 1 \\1\end{array} \right) e ^ { - 3 t }
B) X=c1(21)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { l } 2 \\1\end{array} \right) e ^ { - 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { - 3 t }
C) X=c1(21)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { c } - 2 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
D) X=c1(12)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { c } 1 \\- 2\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
E) X=c1(21)e2t+c2(11)e3tX = c _ { 1 } \left( \begin{array} { c } 2 \\- 1\end{array} \right) e ^ { 2 t } + c _ { 2 } \left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { 3 t }
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29
A particular solution of X=(1211)X+(2t)\mathbf { X } ^ { \prime } = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) \mathbf { X } + \left( \begin{array} { l } 2 \\t\end{array} \right) is

A) Xp=(2t+2t+3)\mathrm { X } _ { \boldsymbol { p } } = \left( \begin{array} { c } 2 t + 2 \\- t + 3\end{array} \right)
B) Xp=(2t+2t+3)\mathrm { X } _ { p } = \left( \begin{array} { c } - 2 t + 2 \\- t + 3\end{array} \right)
C) Xp=(2t+2t3)X _ { p } = \left( \begin{array} { c } - 2 t + 2 \\- t - 3\end{array} \right)
D) Xy=(2t+2t+3)\mathrm { X } _ { y } = \left( \begin{array} { c } - 2 t + 2 \\t + 3\end{array} \right)
E) Xy=(2t2t3)X _ { y } = \left( \begin{array} { c } - 2 t - 2 \\- t - 3\end{array} \right)
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30
Using the improved Euler method with a step size of h=0.1h = 0.1 , the solution of y=1y2,y(0)=0y ^ { \prime } = 1 - y ^ { 2 } , y ( 0 ) = 0 at x=0.1x = 0.1 is

A) y1=0.095y _ { 1 } = 0.095
B) y1=0.995y _ { 1 } = 0.995
C) y1=0.95y _ { 1 } = 0.95
D) y1=0.00995y _ { 1 } = 0.00995
E) y1=0.0995y _ { 1 } = 0.0995
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31
Using the convolution theorem, we find that L1{1/((s+1)(s2+1))}=\mathcal { L } ^ { - 1 } \left\{ 1 / \left( ( s + 1 ) \left( s ^ { 2 } + 1 \right) \right) \right\} =

A) (et+sintcost)/2\left( e ^ { - t } + \sin t - \cos t \right) / 2
B) (et+sintcost)/2\left( e ^ { t } + \sin t - \cos t \right) / 2
C) (et+sint+cost)/2\left( e ^ { - t } + \sin t + \cos t \right) / 2
D) (etsintcost)/2\left( e ^ { t } - \sin t - \cos t \right) / 2
E) (etsintcost)/2\left( e ^ { - t } - \sin t - \cos t \right) / 2
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32
The solution of X=(1211)X\mathbf { X } ^ { \prime } = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) \mathbf { X } is

A) X=c1[(21)cost(01)sint]+c2[(21)sint+(01)cost]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \cos t - \left( \begin{array} { c } 0 \\- 1\end{array} \right) \sin t \right] + c _ { 2 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \sin t + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos t \right]
B) X=c1(10)e3t+c2(01)e3tX = c _ { 1 } \left( \begin{array} { l } 1 \\0\end{array} \right) e ^ { \sqrt { 3 } t } + c _ { 2 } \left( \begin{array} { l } 0 \\1\end{array} \right) e ^ { - \sqrt { 3 } t }
C) X=c1(10)et+c2(01)et\mathbf { X } = c _ { 1 } \left( \begin{array} { l } 1 \\0\end{array} \right) e ^ { t } + c _ { 2 } \left( \begin{array} { l } 0 \\1\end{array} \right) e ^ { - t }
D) X=c1[(21)cos(3t)(01)sin(3t)]+c2[(21)sin(3t)+(01)cos(3t)]\begin{array} { l } \mathrm { X } = c _ { 1 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \cos ( \sqrt { 3 } t ) - \left( \begin{array} { c } 0 \\- 1\end{array} \right) \sin ( \sqrt { 3 } t ) \right] + \\c _ { 2 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \sin ( \sqrt { 3 } t ) + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos ( \sqrt { 3 } t ) \right]\end{array}
E) X=c1[(21)cos(2t)(01)sin(2t)]+c2[(21)sin(2t)+(01)cos(2t)]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \cos ( 2 t ) - \left( \begin{array} { c } 0 \\- 1\end{array} \right) \sin ( 2 t ) \right] + c _ { 2 } \left[ \left( \begin{array} { l } 2 \\1\end{array} \right) \sin ( 2 t ) + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos ( 2 t ) \right]
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33
In the previous two problems, the error in the improved Euler method at x=0.1x = 0.1 is

A) 0.00467
B) 0.000168
C) 0.870
D) 0.895
E) 0.0897
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34
Using Laplace transform methods, the solution of y+y=δ(tπ),y(0)=1,y(0)=0y ^ { \prime \prime } + y = \delta ( t - \pi ) , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is

A) y=sint+sin(tπ)u(tπ)y = \sin t + \sin ( t - \pi ) \boldsymbol { u } ( t - \pi )
B) y=sintcos(tπ)u(tπ)y = \sin t - \cos ( t - \pi ) u ( t - \pi )
C) y=cost+sin(tπ)u(tπ)y = \cos t + \sin ( t - \pi ) u ( t - \pi )
D) y=cost+cos(tπ)u(tπ)y = \cos t + \cos ( t - \pi ) u ( t - \pi )
E) y=costsin(tπ)u(tπ)y = \cos t - \sin ( t - \pi ) u ( t - \pi )
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35
A uniform beam of length 10 has a concentrated load w0w _ { 0 } at x=5x = 5 . It is embedded at both ends. The boundary value problem for the deflections, y(x)y ( x ) , for this system is

A) y=ELw0δ(x5),y(0)=0,y(0)=0,y(10)=0,y(10)=0y ^ { \prime \prime \prime \prime } = E L w _ { 0 } \delta ( x - 5 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
B) y=Elw0δ(x10),y(0)=0,y(0)=0,y(10)=0,y(10)=0y ^ { \prime \prime } = E l w _ { 0 } \delta ( x - 10 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
C) Ely=w0δ(x5),y(0)=0,y(0)=0,y(10)=0,y(10)=0E l y ^ { \prime \prime } = w _ { 0 } \delta ( x - 5 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
D) ELy=w0δ(x5),y(0)=0,y(0)=0,y(10)=0,y(10)=0E L y ^ { \prime \prime \prime } = w _ { 0 } \delta ( x - 5 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
E) ELy=w0δ(x10),y(0)=0,y(0)=0,y(10)=0,y(10)=0E L y ^ { \prime \prime \prime \prime } = w _ { 0 } \delta ( x - 10 ) , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 10 ) = 0 , y ^ { \prime } ( 10 ) = 0
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36
The eigenvalues of the matrix A=(300031011)A = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) are

A) 1, 2, 3
B) 2, 2, 3
C) 1, 2, 2
D) 2,2,3- 2 , - 2,3
E) 1,2,3- 1 , - 2,3
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37
Using power series methods, the solution of 2xy+y+2y=02 x y ^ { \prime \prime } + y ^ { \prime } + 2 y = 0 is

A) y=c0n1(2)nxn/(n!(13(2n1)))+c1x1/2n1(2)nxn/(n!(35(2n+1)))\begin{array} { l } y = c _ { 0 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) + \\c _ { 1 } x ^ { 1 / 2 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) )\end{array}
B) y=c0n1(2)nxn/(n!(13(2n1)))+c1x1/2[1+n1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = c _ { 0 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) + \\c _ { 1 } x ^ { 1 / 2 } \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
C) y=c0[1+n=1(2)nxn/(n!(13(2n1)))]+c1[1+n=1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = c _ { 0 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\c _ { 1 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
D) y=c0[1+n=1(2)nxn/(n!(13(2n1)))]+c1x1/2[1+n=1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = c _ { 0 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\c _ { 1 } x ^ { 1 / 2 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
E) y=[1+n1(2)nxn/(n!(13(2n1)))]+x1/2[1+n1(2)nxn/(n!(35(2n+1)))]\begin{array} { l } y = \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\x ^ { 1 / 2 } \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]\end{array}
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38
Using power series methods, the solution of xyxy+y=0x y ^ { \prime \prime } - x y ^ { \prime } + y = 0 is

A) y=c0x+c1[xlnx1+n=2xn/n!]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n = 2 } ^ { \infty } x ^ { n } / n ! \right]
B) y=c0x+c1[xlnx1+n1xn/(n!(n+1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n - 1 } ^ { \infty } x ^ { n } / ( n ! ( n + 1 ) ) \right]
C) y=c0x+c1[xlnx+n2xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x + \sum _ { n - 2 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
D) y=c0x+c1[xlnx+n1xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x + \sum _ { n - 1 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
E) y=c0x+c1[xlnx1+n2xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n - 2 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
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39
The solution of the eigenvalue problem y+λy=0,y(0)=0,y(1)=0y ^ { \prime \prime } + \lambda y = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 1 ) = 0 is

A) λ=n2π2/4,y=cos(nπx/2),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } / 4 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots
B) λ=nπ/2,y=cos(nπx/2),n=1,2,3,\lambda = n \pi / 2 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots
C) λ=n2π2/4,y=sin(nπx/2),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } / 4 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
D) λ=(2n1)π/2,y=cos((2n1)πx/2),n=1,2,3,\lambda = ( 2 n - 1 ) \pi / 2 , y = \cos ( ( 2 n - 1 ) \pi x / 2 ) , n = 1,2,3 , \ldots
E) λ=(2n1)2π2/4,y=cos((2n1)πx/2),n=1,2,3,\lambda = ( 2 n - 1 ) ^ { 2 } \pi ^ { 2 } / 4 , y = \cos ( ( 2 n - 1 ) \pi x / 2 ) , n = 1,2,3 , \ldots
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40
The eigenvalues of the matrix A=(1211)A = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) are

A) ±3\pm \sqrt { 3 }
B) ±3i\pm \sqrt { 3 } i
C) ±1\pm 1
D) ±2i\pm 2 i
E) ±\pm
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41
In the previous problem, the error in the classical Runge-Kutta method at x=0.1x = 0.1 is (Hint: see the previous five problems.)

A) 0.0008
B) 0.00008
C) 0.00000008
D) 0.000008
E) 0.0000008
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42
The eigenvalues of the matrix A=(300031011)A = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) are

A) (100)\left( \begin{array} { l } 1 \\0 \\0\end{array} \right)
B) (011)\left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right)
C) (110)\left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right)
D) (011)\left( \begin{array} { l } 0 \\1 \\1\end{array} \right)
E) (010)\left( \begin{array} { l } 0 \\1 \\0\end{array} \right)
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43
The differential equation y=x2y2y ^ { \prime } = x ^ { 2 } y ^ { 2 } is Select all that apply.

A) linear
B) separable
C) exact
D) non-linear
E) Bernoulli
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44
Consider the boundary-value problem y4y+3y=x,y(0)=1,y(1)=2y ^ { \prime \prime } - 4 y ^ { \prime } + 3 y = x , y ( 0 ) = 1 , y ( 1 ) = 2 . Replace the derivatives with central differences with a step size of h=1/4h = 1 / 4 . The resulting equations are

A) 12yi+129yi24yi1=xi12 y _ { i + 1 } - 29 y _ { i } - 24 y _ { i - 1 } = x _ { i }
B) 12yi+1+17yi+12yi1=xi12 y _ { i + 1 } + 17 y _ { i } + 12 y _ { i - 1 } = x _ { i }
C) 8yi+1+17yi+12yi1=xi8 y _ { i + 1 } + 17 y _ { i } + 12 y _ { i - 1 } = x _ { i }
D) 8yi+129yi+24yi1=xi8 y _ { i + 1 } - 29 y _ { i } + 24 y _ { i - 1 } = x _ { i }
E) 8yi+1+29yi24yi1=xi8 y _ { i + 1 } + 29 y _ { i } - 24 y _ { i - 1 } = x _ { i }
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45
Using the classical Runge-Kutta method of order 4 with a step size of h=0.1h = 0.1 , the solution of y=1y2,y(0)=0y ^ { \prime } = 1 - y ^ { 2 } , y ( 0 ) = 0 at x=0.1x = 0.1 is

A) 0.099588
B) 0.099668
C) 0.099688
D) 0.099768
E) 0.099788
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46
The differential equation xdyydx=0x d y - y d x = 0 is Select all that apply.

A) linear
B) separable
C) exact
D) non-linear
E) Bernoulli
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47
The differential equation y+y=x2y ^ { \prime } + y = x ^ { 2 } is Select all that apply.

A) linear
B) separable
C) exact
D) non-linear
E) Bernoulli
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48
The solution of the system in the previous problem is

A) y1=1.228,y2=1.482,y3=1.753y _ { 1 } = 1.228 , y _ { 2 } = 1.482 , y _ { 3 } = 1.753
B) y1=1.228,y2=1.646,y3=1.753y _ { 1 } = 1.228 , y _ { 2 } = 1.646 , y _ { 3 } = 1.753
C) y1=1.126,y2=1.646,y3=2.903y _ { 1 } = 1.126 , y _ { 2 } = 1.646 , y _ { 3 } = 2.903
D) y1=1.126,y2=1.786,y3=2.903y _ { 1 } = 1.126 , y _ { 2 } = 1.786 , y _ { 3 } = 2.903
E) y1=1.016,y2=1.786,y3=2.903y _ { 1 } = 1.016 , y _ { 2 } = 1.786 , y _ { 3 } = 2.903
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