Deck 17: Mathematical Problems and Solutions

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Question
The solution of y6y+8y=0y ^ { \prime \prime } - 6 y ^ { \prime } + 8 y = 0 is

A) y=c1e2x+c2e4xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 4 x }
B) y=c1e2x+c2xe4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } x e ^ { 4 x }
C) y=c1e2x+c2xe4xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } x e ^ { - 4 x }
D) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
E) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { - 4 x }
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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)e3tX = 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(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 }
E) 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 }
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
Using power series methods, the solution of 2xy+y+2y=02 x y ^ { \prime \prime } + y ^ { \prime } + 2 y = 0 is

A) y=c0n=1(2)nxn/(n!(13(2n1)))+c1x1/2n=1(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=c0n=1(2)nxn/(n!(13(2n1)))+c1x1/2[1+n=1(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+n=1(2)nxn/(n!(13(2n1)))]+x1/2[1+n=1(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
The solution of y+2y=x+exy ^ { \prime \prime } + 2 y ^ { \prime } = x + e ^ { x } is

A) y=c1+c2e2x+x2/4x/4+ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } + x ^ { 2 } / 4 - x / 4 + e ^ { x } / 3
B) y=c1+c2e2x+x2/4+x/4ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } + x ^ { 2 } / 4 + x / 4 - e ^ { x } / 3
C) y=c1+c2e2x+x2/4+x/4+ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } + x ^ { 2 } / 4 + x / 4 + e ^ { x } / 3
D) y=c1+c2e2xx2/4x/4ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } - x ^ { 2 } / 4 - x / 4 - e ^ { x } / 3
E) y=c1+c2e2xx2/4x/4+ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } - x ^ { 2 } / 4 - x / 4 + e ^ { x } / 3
Question
The solution of xy=(x1)y2x y ^ { \prime } = ( x - 1 ) y ^ { 2 } is

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/(xlnx+c)y = - 1 / ( x - \ln x + c )
Question
In the previous problem, the solution for the temperature is

A) T(t)=7038e.930tT ( t ) = 70 - 38 e ^ { - .930 t }
B) T(t)=7038e.930tT ( t ) = 70 - 38 e ^ { .930 t }
C) T(t)=5532e.930tT ( t ) = 55 - 32 e ^ { - .930 t }
D) T(t)=5532e.930tT ( t ) = 55 - 32 e ^ { .930 t }
E) T(t)=55e.930tT ( t ) = 55 e ^ { - .930 t }
Question
The solution of x2yxy=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+c2x2y = c _ { 1 } + c _ { 2 } x ^ { 2 }
D) y=c1+c2lnxy = c _ { 1 } + c _ { 2 } \ln x
E) y=c1+c2x2y = c _ { 1 } + c _ { 2 } x ^ { - 2 }
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
In the previous problem, the solution for the position, x(t)x ( t ) , is

A) x=e4t(cos(4t)+sin(4t))/2x = e ^ { 4 t } ( \cos ( 4 t ) + \sin ( 4 t ) ) / 2
B) x=e4t(cos(4t)+sin(4t))/2x = e ^ { - 4 t } ( \cos ( 4 t ) + \sin ( 4 t ) ) / 2
C) x=e4t(cos(8t)+sin(8t))/2x = e ^ { - 4 t } ( \cos ( 8 t ) + \sin ( 8 t ) ) / 2
D) x=e4t(cos(8t)+sin(8t))/2x = e ^ { 4 t } ( \cos ( 8 t ) + \sin ( 8 t ) ) / 2
E) x=4e8t32te8tx = 4 e ^ { 8 t } - 32 t e ^ { 8 t }
Question
Using Laplace transform methods, the solution of y+y=δ(tπ/2),y(0)=1y ^ { \prime \prime } + y = \delta ( t - \pi / 2 ) , y ( 0 ) = 1 , y(0)=0y ^ { \prime } ( 0 ) = 0 is

A) y=sint+sin(tπ/2)u(tπ/2)y = \sin t + \sin ( t - \pi / 2 ) u ( t - \pi / 2 )
B) y=sintcos(tπ/2)u(tπ/2)y = \sin t - \cos ( t - \pi / 2 ) \boldsymbol { u } ( t - \pi / 2 )
C) y=cost+sin(tπ/2)u(tπ/2)y = \cos t + \sin ( t - \pi / 2 ) u ( t - \pi / 2 )
D) y=cost+cos(tπ/2)u(tπ/2)y = \cos t + \cos ( t - \pi / 2 ) \boldsymbol { u } ( t - \pi / 2 )
E) y=costsin(tπ/2)u(tπ/2)y = \cos t - \sin ( t - \pi / 2 ) u ( t - \pi / 2 )
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 y+y=xy ^ { \prime } + y = x is

A) y=x+1+cexy = - x + 1 + c e ^ { x }
B) y=x1+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
A 4-pound weight is hung on a spring and stretches it 1 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 6 inches from equilibrium and released, the initial value problem describing the position, x(t)x ( t ) , of the mass at time t is

A) x8x+32x=0,x(0)=6,x(0)=0x ^ { \prime \prime } - 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 6 , x ^ { \prime } ( 0 ) = 0
B) x+8x+32x=0,x(0)=6,x(0)=0x ^ { \prime \prime } + 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 6 , x ^ { \prime } ( 0 ) = 0
C) x8x+32x=0,x(0)=1/2,x(0)=0x ^ { \prime \prime } - 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 1 / 2 , x ^ { \prime } ( 0 ) = 0
D) x+8x+32x=0,x(0)=1/2,x(0)=0x ^ { \prime \prime } + 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 1 / 2 , x ^ { \prime } ( 0 ) = 0
E) x+32x=8,x(0)=1/2,x(0)=0x ^ { \prime \prime } + 32 x = 8 , x ( 0 ) = 1 / 2 , x ^ { \prime } ( 0 ) = 0
Question
The solution of X=(400031011)XX ^ { \prime } = \left( \begin{array} { c c c } 4 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) X are

A) X=c1(010)e4t+c2(011)e2t+c3[(011)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 4 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)e4t+c2(010)e2t+c3[(010)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 4 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)e4t+c2(011)e2t+c3[(011)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 4 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)e4t+c2(011)e2t+c3[(011)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 4 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)e4t+c2(110)e2t+c3[(110)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 4 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 y4y+20y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 20 y = 0 is

A) y=c1e2xcos(4x)+c2e2xsin(4x)y = c _ { 1 } e ^ { - 2 x } \cos ( 4 x ) + c _ { 2 } e ^ { - 2 x } \sin ( 4 x )
B) y=c1e2xcos(4x)+c2e2xsin(4x)y = c _ { 1 } e ^ { - 2 x } \cos ( 4 x ) + c _ { 2 } e ^ { 2 x } \sin ( 4 x )
C) y=c1e2xcos(4x)+c2e2xsin(4x)y = c _ { 1 } e ^ { 2 x } \cos ( 4 x ) + c _ { 2 } e ^ { 2 x } \sin ( 4 x )
D) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
E) y=c1cos(4x)+c2sin(4x)y = c _ { 1 } \cos ( 4 x ) + c _ { 2 } \sin ( 4 x )
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+n=1xn/(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+n=2xn/(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+n=1xn/(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+n=2xn/(n!(n1))]y = c _ { 0 } x + c _ { 1 } \left[ x \ln x - 1 + \sum _ { n = 2 } ^ { \infty } x ^ { n } / ( n ! ( n - 1 ) ) \right]
Question
A frozen chicken at 32F32 ^ { \circ } \mathrm { F } is taken out of the freezer and placed on a table at 70F70 ^ { \circ } \mathrm { F } . One hour later the temperature of the chicken is 55F55 ^ { \circ } \mathrm { F } . 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)=32,T(1)=55\frac { d T } { d t } = k T , T ( 0 ) = 32 , T ( 1 ) = 55
B) dTdt=k(T70),T(0)=32,T(1)=55\frac { d T } { d t } = k ( T - 70 ) , T ( 0 ) = 32 , T ( 1 ) = 55
C) dTdt=(T70),T(0)=32,T(1)=55\frac { d T } { d t } = ( T - 70 ) , T ( 0 ) = 32 , T ( 1 ) = 55
D) dTdt=T,T(0)=32,T(1)=55\frac { d T } { d t } = T , T ( 0 ) = 32 , T ( 1 ) = 55
E) dTdt=k(T55),T(0)=32,T(1)=55\frac { d T } { d t } = k ( T - 55 ) , T ( 0 ) = 32 , T ( 1 ) = 55
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 correct form of the particular solution of y+2y+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
In the previous two problems, the error in the improved Euler method at x=0.1x = 0.1 is

A) 0.004670.00467
B) 0.000165
C) 0.870
D) 0.895
E) 0.0897
Question
Consider the problem 2ur2+1rur+1r22uθ2=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 with boundary conditions u(r,0)=0u ( r , 0 ) = 0 , u(r,π)=0,u(1,θ)=f(θ)u ( r , \pi ) = 0 , u ( 1 , \theta ) = f ( \theta ) . Separate variables using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems for R and ΘR \text { and } \Theta are

A) r2R+rR+λR=0,R(0)=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , R ( 0 ) = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
B) r2R+rR+λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
C) r2R+rRλR=0,R(0)=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , R ( 0 ) = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
D) r2R+rRλR=0,R(0) is bounded, Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , R ( 0 ) \text { is bounded, } \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
E) r2R+rRλR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
Question
The solutions of the eigenvalue problem and the other problem from the previous problem are

A) λ=nπ,X=cos(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n \pi , X = \cos ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
B) λ=nπ,X=sin(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n \pi , X = \sin ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
C) λ=n2π2,X=cos(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } , X = \cos ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
D) λ=n2π2,X=sin(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } , X = \sin ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
E) λ=n2π2,X=cos(nπx),Y=sinh(nπy),n=0,1,2,,(Y=y if n=0)\lambda = n ^ { 2 } \pi ^ { 2 } , X = \cos ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 0,1,2 , \ldots , ( Y = y \text { if } n = 0 )
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) 0.00083
B) 0.000083
C) 0.000000083
D) 0.0000083
E) 0.00000083
Question
In the previous problem, the solution for U(α,t)U ( \alpha , t ) is

A) U=u0(1+ekα2t)/αU = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right) / \alpha
B) U=u0(1ekα2t)/αU = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha
C) U=u0(1ekα2t)/αU = u _ { 0 } \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha
D) U=u0(1ekα2t)U = u _ { 0 } \left( 1 - e ^ { k \alpha ^ { 2 } t } \right)
E) U=u0(1+ekα2t)U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right)
Question
The solution of X=(1121)XX ^ { \prime } = \left( \begin{array} { c c } 1 & 1 \\- 2 & - 1\end{array} \right) X is

A) X=c1[(11)cost(01)sint]+c2[(11)sint+(01)cost]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \cos t - \left( \begin{array} { l } 0 \\1\end{array} \right) \sin t \right] + c _ { 2 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \sin t + \left( \begin{array} { l } 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[(11)cos(3t)(01)sin(3t)]+c2[(11)sin(3t)+(01)cos(3t)]\begin{array} { l } X = c _ { 1 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \cos ( \sqrt { 3 } t ) - \left( \begin{array} { l } 0 \\1\end{array} \right) \sin ( \sqrt { 3 } t ) \right] + \\c _ { 2 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \sin ( \sqrt { 3 } t ) + \left( \begin{array} { l } 0 \\1\end{array} \right) \cos ( \sqrt { 3 } t ) \right]\end{array}
E) X=c1[(11)cos(2t)(01)sin(2t)]+c2[(11)sin(2t)+(01)cos(2t)]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { l } 1 \\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 } 1 \\1\end{array} \right) \sin ( 2 t ) + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos ( 2 t ) \right]
Question
Consider the non-linear system x=12xy,y=2xyyx ^ { \prime } = 1 - 2 x y , y ^ { \prime } = 2 x y - y . The linearized system about the one critical point, (1/2,1), is Xt=AX, where A=( 1 / 2,1 ) \text {, is } X ^ { t } = A X \text {, where } A =

A) (2120)\left( \begin{array} { l l } 2 & 1 \\2 & 0\end{array} \right)
B) (2120)\left( \begin{array} { c c } 2 & - 1 \\2 & 0\end{array} \right)
C) (2120)\left( \begin{array} { c c } - 2 & - 1 \\2 & 0\end{array} \right)
D) (2120)\left( \begin{array} { l l } - 2 & 1 \\- 2 & 0\end{array} \right)
E) (2120)\left( \begin{array} { c c } - 2 & - 1 \\- 2 & 0\end{array} \right)
Question
Consider the heat problem k2ux2=ut,0<x<,t>0,u(x,0)=0,u(0,t)=u0k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , 0 < x < \infty , t > 0 , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 } . Apply a Fourier sine transform. The resulting problem for U(α,t)=Fs{u(x,t)}U ( \alpha , t ) = \mathcal { F } _ { s } \{ u ( x , t ) \} is

A) Ut=kαU+kαu0,U(α,0)=0U _ { t } = - k \alpha U + k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0
B) Ut=kα2Ukαt0,U(α,0)=0U _ { t } = - k \alpha ^ { 2 } U - k \alpha t _ { 0 } , U ( \alpha , 0 ) = 0
C) Ut=kα2U+kαt0,U(α,0)=0U _ { t } = - k \alpha ^ { 2 } U + k \alpha t _ { 0 } , U ( \alpha , 0 ) = 0
D) Ut=kα2U+kαu0,U(α,0)=0U _ { t } = k \alpha ^ { 2 } U + k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0
E) Ut=kα2Ukαu0,U(α,0)=0U _ { t } = k \alpha ^ { 2 } U - k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0
Question
Let A=(4948)A = \left( \begin{array} { c c } - 4 & - 9 \\4 & 8\end{array} \right) , and consider the system X=AXX ^ { \prime } = A X . The critical point (0,0)( 0,0 ) of the system is a

A) stable node
B) unstable node
C) unstable saddle
D) stable spiral point
E) unstable spiral point
Question
Consider Laplace's equation on a rectangle, 2ux2+2uy2=0\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = 0 with boundary conditions ux(0,y)=0,ux(1,y)=0,u(x,0)=0,u(x,2)=f(x)u _ { x } ( 0 , y ) = 0 , u _ { x } ( 1 , y ) = 0 , u ( x , 0 ) = 0 , u ( x , 2 ) = f ( x ) . When the variables are separated using u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) , the resulting problems for XX and YY are

A) X+λX=0,X(0)=0,X(1)=0,YλY=0,Y(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( 1 ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 0 ) = 0
B) X+λX=0,X(0)=0,X(1)=0,Y+λY=0,Y(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( 1 ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( 0 ) = 0
C) X+λX=0,X(0)=0,X(1)=0,YλY=0,Y(2)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( 1 ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 2 ) = 0
D) X+λX=0,X(0)=0,Y+λY=0,Y(0)=0,Y(2)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( 0 ) = 0 , Y ( 2 ) = 0
E) X+λX=0,X(0)=0,YλY=0,Y(0)=0,Y(2)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 0 ) = 0 , Y ( 2 ) = 0
Question
Using the improved Euler method with a step size of h=0.1h = 0.1 , the solution of y=1+y2,y(0)=0 at x=0.1y ^ { \prime } = 1 + y ^ { 2 } , y ( 0 ) = 0 \text { at } x = 0.1 is

A) y1=0.1015y _ { 1 } = 0.1015
B) y1=0.115y _ { 1 } = 0.115
C) y1=0.105y _ { 1 } = 0.105
D) y1=0.10005y _ { 1 } = 0.10005
E) y1=0.1005y _ { 1 } = 0.1005
Question
The solution of the eigenvalue problem y+λy=0,y(0)=0,y(2)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 2 ) = 0 is

A) 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
B) λ=(nπ/2)2,y=cos(nπx/2),n=1,2,3,\lambda = ( n \pi / 2 ) ^ { 2 } , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots
C) λ2=nπ/2,y=sin(nπx/2),n=1,2,3,\lambda ^ { 2 } = n \pi / 2 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
D) λ=nπ/2,y=sin(nπx/2),n=1,2,3,\lambda = n \pi / 2 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
E) λ=(nπ/2)2,y=sin(nπx/2),n=1,2,3,\lambda = ( n \pi / 2 ) ^ { 2 } , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
Question
In the previous problem, the exact solution of the initial value problem is

A) y=tanxy = \tan x
B) y=secxy = \sec x
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
In the previous problem, for both the linearized system and the non-linear system, the critical point is a

A) unstable node
B) stable node
C) saddle point
D) unstable spiral point
E) stable spiral point
Question
In the previous two problems, the solution for u(x,y)u ( x , y ) is

A) u=n=1cncos(nπx)sinh(nπx), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \sinh ( n \pi x ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
B) u=n=1cnsin(nπx)sinh(nπy), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x ) \sinh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
C) u=c0y+n=1cncos(nπx)sinh(nπy)u = c _ { 0 } y + \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \sinh ( n \pi y ) , where cn=02f(x)dx/4c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) d x / 4 and
cn=02f(x)cos(nπx)dx/sinh(2nπ)c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
D) u=n=1cncos(nπx)cosh(nπy), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \cosh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
E) u=n=1cnsin(nπx)cosh(nπy), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x ) \cosh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
Question
A particular solution of X=(1121)X+(2t)\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & 1 \\- 2 & - 1\end{array} \right) \mathbf { X } + \left( \begin{array} { l } 2 \\t\end{array} \right) is

A) Xp=(t+2t+3)X _ { p } = \left( \begin{array} { c } t + 2 \\- t + 3\end{array} \right)
B) Xp=(t+2t3)X _ { p } = \left( \begin{array} { c } t + 2 \\- t - 3\end{array} \right)
C) Xp=(t+2t3)X _ { p } = \left( \begin{array} { l } - t + 2 \\- t - 3\end{array} \right)
D) Xp=(t+2t+3)\mathbf { X } _ { p } = \left( \begin{array} { c } - t + 2 \\t + 3\end{array} \right)
E) Xp=(t2t+3)X _ { p } = \left( \begin{array} { l } - t - 2 \\- t + 3\end{array} \right)
Question
Let A=(1411)A = \left( \begin{array} { c c } - 1 & - 4 \\1 & - 1\end{array} \right) , and consider the system X=AXX ^ { \prime } = A X . The critical point (0,0)( 0,0 ) of the system is a spiral point. The origin is

A) unstable, and the solutions recede from the origin clockwise as tt \rightarrow \infty .
B) unstable, and the solutions recede from the origin counter-clockwise as tt \rightarrow \infty .
C) stable, and the solutions approach the origin clockwise as tt \rightarrow \infty .
D) stable, and the solutions approach the origin counter-clockwise as tt \rightarrow \infty .
E) none of the above
Question
In the previous two problems, the infinite series solution for u(r,θ)u ( r , \theta ) is u=n=1cnrnΘn(θ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } r ^ { n } \Theta _ { n } ( \theta ) , where Θn\Theta _ { n } is found in the previous problem, and

A) cn=20πf(θ)sin(nθ)dθ/πc _ { n } = 2 \int _ { 0 } ^ { \pi } f ( \theta ) \sin ( n \theta ) d \theta / \pi
B) cn=20πf(θ)cos(nθ)dθ/πc _ { n } = 2 \int _ { 0 } ^ { \pi } f ( \theta ) \cos ( n \theta ) d \theta / \pi
C) cn=0πf(θ)cos(nθ)dθ/πc _ { n } = \int _ { 0 } ^ { \pi } f ( \theta ) \cos ( n \theta ) d \theta / \pi
D) cn=0πf(θ)sin(nθ)dθ/πc _ { n } = \int _ { 0 } ^ { \pi } f ( \theta ) \sin ( n \theta ) d \theta / \pi
E) cn=0πf(θ)sin(nθ)dθ/(2π)c _ { n } = \int _ { 0 } ^ { \pi } f ( \theta ) \sin ( n \theta ) d \theta / ( 2 \pi )
Question
The solutions for λ,R and Θ\lambda , R \text { and } \Theta from the previous problem are

A) λ=n2,R=rn,Θ=cos(nθ),n=1,2,3,\lambda = n ^ { 2 } , R = r ^ { n } , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
B) λ=n2,R=rn,Θ=sin(nθ),n=1,2,3,\lambda = n ^ { 2 } , R = r ^ { n } , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
C) λ=n2,R=rn,Θ=sin(nθ),n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , \Theta = \sin ( n \theta ) , n = 0,1,2 , \ldots
D) λ=n,R=rn,Θ=sin(nθ),n=1,2,3,\lambda = n , R = r ^ { n } , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
E) λ=n,R=rn,Θ=cos(nθ),n=1,2,3,\lambda = n , R = r ^ { n } , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
Question
Using the classical Runge-Kutta method of order 4 with a step size of h=0.1h = 0.1 , the solution of y=1+y2,y(0) at x=0.1y ^ { \prime } = 1 + y ^ { 2 } , y ( 0 ) \text { at } x = 0.1 is

A) 0.099589
B) 0.100334589
C) 0.10034589
D) 0.10334589
E) 0.1034589
Question
In the previous two problems, the solution for u along the line t=0.5t = 0.5 at the mesh points is Select all that apply.

A) u11=10/3u _ { 11 } = 10 / 3
B) u11=20/9u _ { 11 } = 20 / 9
C) u11=20/3u _ { 11 } = 20 / 3
D) u21=32/3u _ { 21 } = 32 / 3
E) u21=13/3u _ { 21 } = 13 / 3
Question
In the previous problem, using the notation uij=u(x,t)u _ { i j } = u ( x , t ) , and letting c=1,λ=ck/h2c = 1 , \lambda = c k / h ^ { 2 } , the equation becomes

A) ui,j1=λui+1,j+(1+2λ)ui,j+λui1,ju _ { i , j - 1 } = \lambda u _ { i + 1 , j } + ( 1 + 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
B) ui,j1=λui+1,j+(12λ)ui,j+λui1,ju _ { i , j - 1 } = \lambda u _ { i + 1 , j } + ( 1 - 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
C) ui,j+1=λui+1,j+(1+2λ)ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 + 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
D) ui,j+1=λui+1,j+(12λ)ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
E) ui,j+1=λui+1,j+(1λ)ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
Question
Is the value of λ\lambda in the previous problem such that the scheme is stable?

A) yes
B) no
C) It is right on the borderline.
D) It cannot be determined from the available data.
Question
The Fourier series of an even function can contain Select all that apply.

A) sine terms
B) cosine terms
C) a constant term
D) more than one of the above
E) none of the above
Question
The eigenvalue-eigenvector pairs for the matrix A=(400031011)XA = \left( \begin{array} { c c c } 4 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X } are Select all that apply.

A) 4,(100)4 , \left( \begin{array} { l } 1 \\0 \\0\end{array} \right)
B) 2,(011)2 , \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right)
C) 2,(110)2 , \left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right)
D) 2,(011)2 , \left( \begin{array} { l } 0 \\1 \\1\end{array} \right)
E) 2,(010)2 , \left( \begin{array} { l } 0 \\1 \\0\end{array} \right)
Question
In the previous two problem, the solution for u(x,t)u ( x , t ) is

A) u=2u00[(1ekα2t)sin(αx)/α]dα/πu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha / \pi
B) u=2u00[(1ekα2t)sin(αx)/α]dαu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha
C) u=u00[(1ekα2t)sin(αx)/α]dαu = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha
D) u=u00[(1ekα2t)sin(αx)/α]dα/πu = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha / \pi
E) u=u00[(1ekα2t)sin(x)/α]dα/(2π)u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \infty x ) / \alpha \right] d \alpha / ( 2 \pi )
Question
The solutions of a regular Sturm-Liouville problem ((ry)+(λp+q)y=0,y(a)=0,y(b)=0)\left( \left( r y ^ { \prime } \right) ^ { \prime } + ( \lambda p + q ) y = 0 , y ( a ) = 0 , y ( b ) = 0 \right) have which of the following properties?

A) There exists an infinite number of real eigenvalues.
B) The eigenvalues are orthogonal on [a,b][ a , b ] .
C) For each eigenvalue, there is only one eigenfunction (except for non-zero constant multiples).
D) Eigenfunctions corresponding to different eigenvalues are linearly independent.
E) The set of eigenfunctions corresponding to the set of eigenvalues is orthogonal with respect to the weight function r(x)r ( x ) on the interval [a,b][ a , b ] .
Question
Consider the heat problem c2ux2=ut,u(0,t)=0,u(1,t)=3,u(x,0)=3x2c \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , u ( 1 , t ) = 3 , u ( x , 0 ) = 3 x ^ { 2 } . Replace 2ux2\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } with a central difference approximation with h=1/3h = 1 / 3 and ut\frac { \partial u } { \partial t } with a forward difference approximation with k=1/2k = 1 / 2 . The resulting equation is

A) c[u(x+h,t)+2u(x,t)+u(xh,t)]/h2=(u(x,t+k)u(x,t))/kc [ u ( x + h , t ) + 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k
B) c[u(x+h,t)+2u(x,t)+u(xh,t)]/h2=(u(x,t+k)+u(x,t))/kc [ u ( x + h , t ) + 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) + u ( x , t ) ) / k
C) c[u(x+h,t)2u(x,t)+u(xh,t)]/h2=(u(x,t+k)u(x,t))/kc [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k
D) c[u(x+h,t)4u(x,t)+u(xh,t)]/h2=(u(x,t+k)+u(x,t))/kc [ u ( x + h , t ) - 4 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) + u ( x , t ) ) / k
E) c[u(x+h,t)4u(x,t)+u(xh,t)]/h2=(u(x,t+k)u(x,t))/kc [ u ( x + h , t ) - 4 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k
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Deck 17: Mathematical Problems and Solutions
1
The solution of y6y+8y=0y ^ { \prime \prime } - 6 y ^ { \prime } + 8 y = 0 is

A) y=c1e2x+c2e4xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } e ^ { - 4 x }
B) y=c1e2x+c2xe4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } x e ^ { 4 x }
C) y=c1e2x+c2xe4xy = c _ { 1 } e ^ { - 2 x } + c _ { 2 } x e ^ { - 4 x }
D) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
E) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { - 4 x }
y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
2
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)e3tX = 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(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 }
E) 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 }
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 }
3
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
(et+sintcost)/2\left( e ^ { - t } + \sin t - \cos t \right) / 2
4
Using power series methods, the solution of 2xy+y+2y=02 x y ^ { \prime \prime } + y ^ { \prime } + 2 y = 0 is

A) y=c0n=1(2)nxn/(n!(13(2n1)))+c1x1/2n=1(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=c0n=1(2)nxn/(n!(13(2n1)))+c1x1/2[1+n=1(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+n=1(2)nxn/(n!(13(2n1)))]+x1/2[1+n=1(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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5
The solution of y+2y=x+exy ^ { \prime \prime } + 2 y ^ { \prime } = x + e ^ { x } is

A) y=c1+c2e2x+x2/4x/4+ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } + x ^ { 2 } / 4 - x / 4 + e ^ { x } / 3
B) y=c1+c2e2x+x2/4+x/4ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } + x ^ { 2 } / 4 + x / 4 - e ^ { x } / 3
C) y=c1+c2e2x+x2/4+x/4+ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } + x ^ { 2 } / 4 + x / 4 + e ^ { x } / 3
D) y=c1+c2e2xx2/4x/4ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } - x ^ { 2 } / 4 - x / 4 - e ^ { x } / 3
E) y=c1+c2e2xx2/4x/4+ex/3y = c _ { 1 } + c _ { 2 } e ^ { - 2 x } - x ^ { 2 } / 4 - x / 4 + e ^ { x } / 3
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6
The solution of xy=(x1)y2x y ^ { \prime } = ( x - 1 ) y ^ { 2 } is

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/(xlnx+c)y = - 1 / ( x - \ln x + c )
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7
In the previous problem, the solution for the temperature is

A) T(t)=7038e.930tT ( t ) = 70 - 38 e ^ { - .930 t }
B) T(t)=7038e.930tT ( t ) = 70 - 38 e ^ { .930 t }
C) T(t)=5532e.930tT ( t ) = 55 - 32 e ^ { - .930 t }
D) T(t)=5532e.930tT ( t ) = 55 - 32 e ^ { .930 t }
E) T(t)=55e.930tT ( t ) = 55 e ^ { - .930 t }
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8
The solution of x2yxy=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+c2x2y = c _ { 1 } + c _ { 2 } x ^ { 2 }
D) y=c1+c2lnxy = c _ { 1 } + c _ { 2 } \ln x
E) y=c1+c2x2y = c _ { 1 } + c _ { 2 } x ^ { - 2 }
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9
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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10
In the previous problem, the solution for the position, x(t)x ( t ) , is

A) x=e4t(cos(4t)+sin(4t))/2x = e ^ { 4 t } ( \cos ( 4 t ) + \sin ( 4 t ) ) / 2
B) x=e4t(cos(4t)+sin(4t))/2x = e ^ { - 4 t } ( \cos ( 4 t ) + \sin ( 4 t ) ) / 2
C) x=e4t(cos(8t)+sin(8t))/2x = e ^ { - 4 t } ( \cos ( 8 t ) + \sin ( 8 t ) ) / 2
D) x=e4t(cos(8t)+sin(8t))/2x = e ^ { 4 t } ( \cos ( 8 t ) + \sin ( 8 t ) ) / 2
E) x=4e8t32te8tx = 4 e ^ { 8 t } - 32 t e ^ { 8 t }
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11
Using Laplace transform methods, the solution of y+y=δ(tπ/2),y(0)=1y ^ { \prime \prime } + y = \delta ( t - \pi / 2 ) , y ( 0 ) = 1 , y(0)=0y ^ { \prime } ( 0 ) = 0 is

A) y=sint+sin(tπ/2)u(tπ/2)y = \sin t + \sin ( t - \pi / 2 ) u ( t - \pi / 2 )
B) y=sintcos(tπ/2)u(tπ/2)y = \sin t - \cos ( t - \pi / 2 ) \boldsymbol { u } ( t - \pi / 2 )
C) y=cost+sin(tπ/2)u(tπ/2)y = \cos t + \sin ( t - \pi / 2 ) u ( t - \pi / 2 )
D) y=cost+cos(tπ/2)u(tπ/2)y = \cos t + \cos ( t - \pi / 2 ) \boldsymbol { u } ( t - \pi / 2 )
E) y=costsin(tπ/2)u(tπ/2)y = \cos t - \sin ( t - \pi / 2 ) u ( t - \pi / 2 )
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12
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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13
The solution of y+y=xy ^ { \prime } + y = x is

A) y=x+1+cexy = - x + 1 + c e ^ { x }
B) y=x1+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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14
A 4-pound weight is hung on a spring and stretches it 1 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 6 inches from equilibrium and released, the initial value problem describing the position, x(t)x ( t ) , of the mass at time t is

A) x8x+32x=0,x(0)=6,x(0)=0x ^ { \prime \prime } - 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 6 , x ^ { \prime } ( 0 ) = 0
B) x+8x+32x=0,x(0)=6,x(0)=0x ^ { \prime \prime } + 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 6 , x ^ { \prime } ( 0 ) = 0
C) x8x+32x=0,x(0)=1/2,x(0)=0x ^ { \prime \prime } - 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 1 / 2 , x ^ { \prime } ( 0 ) = 0
D) x+8x+32x=0,x(0)=1/2,x(0)=0x ^ { \prime \prime } + 8 x ^ { \prime } + 32 x = 0 , x ( 0 ) = 1 / 2 , x ^ { \prime } ( 0 ) = 0
E) x+32x=8,x(0)=1/2,x(0)=0x ^ { \prime \prime } + 32 x = 8 , x ( 0 ) = 1 / 2 , x ^ { \prime } ( 0 ) = 0
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15
The solution of X=(400031011)XX ^ { \prime } = \left( \begin{array} { c c c } 4 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) X are

A) X=c1(010)e4t+c2(011)e2t+c3[(011)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 0 \\1 \\0\end{array} \right) e ^ { 4 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)e4t+c2(010)e2t+c3[(010)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right) e ^ { 4 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)e4t+c2(011)e2t+c3[(011)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 4 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)e4t+c2(011)e2t+c3[(011)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 4 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)e4t+c2(110)e2t+c3[(110)te2t+(010)e2t]X = c _ { 1 } \left( \begin{array} { l } 1 \\0 \\0\end{array} \right) e ^ { 4 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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16
The solution of y4y+20y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 20 y = 0 is

A) y=c1e2xcos(4x)+c2e2xsin(4x)y = c _ { 1 } e ^ { - 2 x } \cos ( 4 x ) + c _ { 2 } e ^ { - 2 x } \sin ( 4 x )
B) y=c1e2xcos(4x)+c2e2xsin(4x)y = c _ { 1 } e ^ { - 2 x } \cos ( 4 x ) + c _ { 2 } e ^ { 2 x } \sin ( 4 x )
C) y=c1e2xcos(4x)+c2e2xsin(4x)y = c _ { 1 } e ^ { 2 x } \cos ( 4 x ) + c _ { 2 } e ^ { 2 x } \sin ( 4 x )
D) y=c1e2x+c2e4xy = c _ { 1 } e ^ { 2 x } + c _ { 2 } e ^ { 4 x }
E) y=c1cos(4x)+c2sin(4x)y = c _ { 1 } \cos ( 4 x ) + c _ { 2 } \sin ( 4 x )
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17
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+n=1xn/(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+n=2xn/(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+n=1xn/(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+n=2xn/(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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18
A frozen chicken at 32F32 ^ { \circ } \mathrm { F } is taken out of the freezer and placed on a table at 70F70 ^ { \circ } \mathrm { F } . One hour later the temperature of the chicken is 55F55 ^ { \circ } \mathrm { F } . 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)=32,T(1)=55\frac { d T } { d t } = k T , T ( 0 ) = 32 , T ( 1 ) = 55
B) dTdt=k(T70),T(0)=32,T(1)=55\frac { d T } { d t } = k ( T - 70 ) , T ( 0 ) = 32 , T ( 1 ) = 55
C) dTdt=(T70),T(0)=32,T(1)=55\frac { d T } { d t } = ( T - 70 ) , T ( 0 ) = 32 , T ( 1 ) = 55
D) dTdt=T,T(0)=32,T(1)=55\frac { d T } { d t } = T , T ( 0 ) = 32 , T ( 1 ) = 55
E) dTdt=k(T55),T(0)=32,T(1)=55\frac { d T } { d t } = k ( T - 55 ) , T ( 0 ) = 32 , T ( 1 ) = 55
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19
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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20
The correct form of the particular solution of y+2y+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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21
In the previous two problems, the error in the improved Euler method at x=0.1x = 0.1 is

A) 0.004670.00467
B) 0.000165
C) 0.870
D) 0.895
E) 0.0897
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22
Consider the problem 2ur2+1rur+1r22uθ2=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 with boundary conditions u(r,0)=0u ( r , 0 ) = 0 , u(r,π)=0,u(1,θ)=f(θ)u ( r , \pi ) = 0 , u ( 1 , \theta ) = f ( \theta ) . Separate variables using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems for R and ΘR \text { and } \Theta are

A) r2R+rR+λR=0,R(0)=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , R ( 0 ) = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
B) r2R+rR+λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
C) r2R+rRλR=0,R(0)=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , R ( 0 ) = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
D) r2R+rRλR=0,R(0) is bounded, Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , R ( 0 ) \text { is bounded, } \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
E) r2R+rRλR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
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23
The solutions of the eigenvalue problem and the other problem from the previous problem are

A) λ=nπ,X=cos(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n \pi , X = \cos ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
B) λ=nπ,X=sin(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n \pi , X = \sin ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
C) λ=n2π2,X=cos(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } , X = \cos ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
D) λ=n2π2,X=sin(nπx),Y=sinh(nπy),n=1,2,3,\lambda = n ^ { 2 } \pi ^ { 2 } , X = \sin ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 1,2,3 , \ldots
E) λ=n2π2,X=cos(nπx),Y=sinh(nπy),n=0,1,2,,(Y=y if n=0)\lambda = n ^ { 2 } \pi ^ { 2 } , X = \cos ( n \pi x ) , Y = \sinh ( n \pi y ) , n = 0,1,2 , \ldots , ( Y = y \text { if } n = 0 )
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24
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) 0.00083
B) 0.000083
C) 0.000000083
D) 0.0000083
E) 0.00000083
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25
In the previous problem, the solution for U(α,t)U ( \alpha , t ) is

A) U=u0(1+ekα2t)/αU = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right) / \alpha
B) U=u0(1ekα2t)/αU = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha
C) U=u0(1ekα2t)/αU = u _ { 0 } \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha
D) U=u0(1ekα2t)U = u _ { 0 } \left( 1 - e ^ { k \alpha ^ { 2 } t } \right)
E) U=u0(1+ekα2t)U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right)
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26
The solution of X=(1121)XX ^ { \prime } = \left( \begin{array} { c c } 1 & 1 \\- 2 & - 1\end{array} \right) X is

A) X=c1[(11)cost(01)sint]+c2[(11)sint+(01)cost]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \cos t - \left( \begin{array} { l } 0 \\1\end{array} \right) \sin t \right] + c _ { 2 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \sin t + \left( \begin{array} { l } 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[(11)cos(3t)(01)sin(3t)]+c2[(11)sin(3t)+(01)cos(3t)]\begin{array} { l } X = c _ { 1 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \cos ( \sqrt { 3 } t ) - \left( \begin{array} { l } 0 \\1\end{array} \right) \sin ( \sqrt { 3 } t ) \right] + \\c _ { 2 } \left[ \left( \begin{array} { c } 1 \\- 1\end{array} \right) \sin ( \sqrt { 3 } t ) + \left( \begin{array} { l } 0 \\1\end{array} \right) \cos ( \sqrt { 3 } t ) \right]\end{array}
E) X=c1[(11)cos(2t)(01)sin(2t)]+c2[(11)sin(2t)+(01)cos(2t)]\mathbf { X } = c _ { 1 } \left[ \left( \begin{array} { l } 1 \\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 } 1 \\1\end{array} \right) \sin ( 2 t ) + \left( \begin{array} { c } 0 \\- 1\end{array} \right) \cos ( 2 t ) \right]
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27
Consider the non-linear system x=12xy,y=2xyyx ^ { \prime } = 1 - 2 x y , y ^ { \prime } = 2 x y - y . The linearized system about the one critical point, (1/2,1), is Xt=AX, where A=( 1 / 2,1 ) \text {, is } X ^ { t } = A X \text {, where } A =

A) (2120)\left( \begin{array} { l l } 2 & 1 \\2 & 0\end{array} \right)
B) (2120)\left( \begin{array} { c c } 2 & - 1 \\2 & 0\end{array} \right)
C) (2120)\left( \begin{array} { c c } - 2 & - 1 \\2 & 0\end{array} \right)
D) (2120)\left( \begin{array} { l l } - 2 & 1 \\- 2 & 0\end{array} \right)
E) (2120)\left( \begin{array} { c c } - 2 & - 1 \\- 2 & 0\end{array} \right)
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28
Consider the heat problem k2ux2=ut,0<x<,t>0,u(x,0)=0,u(0,t)=u0k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , 0 < x < \infty , t > 0 , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 } . Apply a Fourier sine transform. The resulting problem for U(α,t)=Fs{u(x,t)}U ( \alpha , t ) = \mathcal { F } _ { s } \{ u ( x , t ) \} is

A) Ut=kαU+kαu0,U(α,0)=0U _ { t } = - k \alpha U + k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0
B) Ut=kα2Ukαt0,U(α,0)=0U _ { t } = - k \alpha ^ { 2 } U - k \alpha t _ { 0 } , U ( \alpha , 0 ) = 0
C) Ut=kα2U+kαt0,U(α,0)=0U _ { t } = - k \alpha ^ { 2 } U + k \alpha t _ { 0 } , U ( \alpha , 0 ) = 0
D) Ut=kα2U+kαu0,U(α,0)=0U _ { t } = k \alpha ^ { 2 } U + k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0
E) Ut=kα2Ukαu0,U(α,0)=0U _ { t } = k \alpha ^ { 2 } U - k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0
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29
Let A=(4948)A = \left( \begin{array} { c c } - 4 & - 9 \\4 & 8\end{array} \right) , and consider the system X=AXX ^ { \prime } = A X . The critical point (0,0)( 0,0 ) of the system is a

A) stable node
B) unstable node
C) unstable saddle
D) stable spiral point
E) unstable spiral point
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30
Consider Laplace's equation on a rectangle, 2ux2+2uy2=0\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = 0 with boundary conditions ux(0,y)=0,ux(1,y)=0,u(x,0)=0,u(x,2)=f(x)u _ { x } ( 0 , y ) = 0 , u _ { x } ( 1 , y ) = 0 , u ( x , 0 ) = 0 , u ( x , 2 ) = f ( x ) . When the variables are separated using u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) , the resulting problems for XX and YY are

A) X+λX=0,X(0)=0,X(1)=0,YλY=0,Y(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( 1 ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 0 ) = 0
B) X+λX=0,X(0)=0,X(1)=0,Y+λY=0,Y(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( 1 ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( 0 ) = 0
C) X+λX=0,X(0)=0,X(1)=0,YλY=0,Y(2)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( 1 ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 2 ) = 0
D) X+λX=0,X(0)=0,Y+λY=0,Y(0)=0,Y(2)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( 0 ) = 0 , Y ( 2 ) = 0
E) X+λX=0,X(0)=0,YλY=0,Y(0)=0,Y(2)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 0 ) = 0 , Y ( 2 ) = 0
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31
Using the improved Euler method with a step size of h=0.1h = 0.1 , the solution of y=1+y2,y(0)=0 at x=0.1y ^ { \prime } = 1 + y ^ { 2 } , y ( 0 ) = 0 \text { at } x = 0.1 is

A) y1=0.1015y _ { 1 } = 0.1015
B) y1=0.115y _ { 1 } = 0.115
C) y1=0.105y _ { 1 } = 0.105
D) y1=0.10005y _ { 1 } = 0.10005
E) y1=0.1005y _ { 1 } = 0.1005
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32
The solution of the eigenvalue problem y+λy=0,y(0)=0,y(2)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 2 ) = 0 is

A) 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
B) λ=(nπ/2)2,y=cos(nπx/2),n=1,2,3,\lambda = ( n \pi / 2 ) ^ { 2 } , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots
C) λ2=nπ/2,y=sin(nπx/2),n=1,2,3,\lambda ^ { 2 } = n \pi / 2 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
D) λ=nπ/2,y=sin(nπx/2),n=1,2,3,\lambda = n \pi / 2 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
E) λ=(nπ/2)2,y=sin(nπx/2),n=1,2,3,\lambda = ( n \pi / 2 ) ^ { 2 } , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots
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33
In the previous problem, the exact solution of the initial value problem is

A) y=tanxy = \tan x
B) y=secxy = \sec x
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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34
In the previous problem, for both the linearized system and the non-linear system, the critical point is a

A) unstable node
B) stable node
C) saddle point
D) unstable spiral point
E) stable spiral point
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35
In the previous two problems, the solution for u(x,y)u ( x , y ) is

A) u=n=1cncos(nπx)sinh(nπx), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \sinh ( n \pi x ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
B) u=n=1cnsin(nπx)sinh(nπy), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x ) \sinh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
C) u=c0y+n=1cncos(nπx)sinh(nπy)u = c _ { 0 } y + \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \sinh ( n \pi y ) , where cn=02f(x)dx/4c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) d x / 4 and
cn=02f(x)cos(nπx)dx/sinh(2nπ)c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
D) u=n=1cncos(nπx)cosh(nπy), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \cosh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
E) u=n=1cnsin(nπx)cosh(nπy), where cn=02f(x)cos(nπx)dx/sinh(2nπ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x ) \cosh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
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36
A particular solution of X=(1121)X+(2t)\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & 1 \\- 2 & - 1\end{array} \right) \mathbf { X } + \left( \begin{array} { l } 2 \\t\end{array} \right) is

A) Xp=(t+2t+3)X _ { p } = \left( \begin{array} { c } t + 2 \\- t + 3\end{array} \right)
B) Xp=(t+2t3)X _ { p } = \left( \begin{array} { c } t + 2 \\- t - 3\end{array} \right)
C) Xp=(t+2t3)X _ { p } = \left( \begin{array} { l } - t + 2 \\- t - 3\end{array} \right)
D) Xp=(t+2t+3)\mathbf { X } _ { p } = \left( \begin{array} { c } - t + 2 \\t + 3\end{array} \right)
E) Xp=(t2t+3)X _ { p } = \left( \begin{array} { l } - t - 2 \\- t + 3\end{array} \right)
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37
Let A=(1411)A = \left( \begin{array} { c c } - 1 & - 4 \\1 & - 1\end{array} \right) , and consider the system X=AXX ^ { \prime } = A X . The critical point (0,0)( 0,0 ) of the system is a spiral point. The origin is

A) unstable, and the solutions recede from the origin clockwise as tt \rightarrow \infty .
B) unstable, and the solutions recede from the origin counter-clockwise as tt \rightarrow \infty .
C) stable, and the solutions approach the origin clockwise as tt \rightarrow \infty .
D) stable, and the solutions approach the origin counter-clockwise as tt \rightarrow \infty .
E) none of the above
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38
In the previous two problems, the infinite series solution for u(r,θ)u ( r , \theta ) is u=n=1cnrnΘn(θ)u = \sum _ { n = 1 } ^ { \infty } c _ { n } r ^ { n } \Theta _ { n } ( \theta ) , where Θn\Theta _ { n } is found in the previous problem, and

A) cn=20πf(θ)sin(nθ)dθ/πc _ { n } = 2 \int _ { 0 } ^ { \pi } f ( \theta ) \sin ( n \theta ) d \theta / \pi
B) cn=20πf(θ)cos(nθ)dθ/πc _ { n } = 2 \int _ { 0 } ^ { \pi } f ( \theta ) \cos ( n \theta ) d \theta / \pi
C) cn=0πf(θ)cos(nθ)dθ/πc _ { n } = \int _ { 0 } ^ { \pi } f ( \theta ) \cos ( n \theta ) d \theta / \pi
D) cn=0πf(θ)sin(nθ)dθ/πc _ { n } = \int _ { 0 } ^ { \pi } f ( \theta ) \sin ( n \theta ) d \theta / \pi
E) cn=0πf(θ)sin(nθ)dθ/(2π)c _ { n } = \int _ { 0 } ^ { \pi } f ( \theta ) \sin ( n \theta ) d \theta / ( 2 \pi )
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39
The solutions for λ,R and Θ\lambda , R \text { and } \Theta from the previous problem are

A) λ=n2,R=rn,Θ=cos(nθ),n=1,2,3,\lambda = n ^ { 2 } , R = r ^ { n } , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
B) λ=n2,R=rn,Θ=sin(nθ),n=1,2,3,\lambda = n ^ { 2 } , R = r ^ { n } , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
C) λ=n2,R=rn,Θ=sin(nθ),n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , \Theta = \sin ( n \theta ) , n = 0,1,2 , \ldots
D) λ=n,R=rn,Θ=sin(nθ),n=1,2,3,\lambda = n , R = r ^ { n } , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
E) λ=n,R=rn,Θ=cos(nθ),n=1,2,3,\lambda = n , R = r ^ { n } , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
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40
Using the classical Runge-Kutta method of order 4 with a step size of h=0.1h = 0.1 , the solution of y=1+y2,y(0) at x=0.1y ^ { \prime } = 1 + y ^ { 2 } , y ( 0 ) \text { at } x = 0.1 is

A) 0.099589
B) 0.100334589
C) 0.10034589
D) 0.10334589
E) 0.1034589
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41
In the previous two problems, the solution for u along the line t=0.5t = 0.5 at the mesh points is Select all that apply.

A) u11=10/3u _ { 11 } = 10 / 3
B) u11=20/9u _ { 11 } = 20 / 9
C) u11=20/3u _ { 11 } = 20 / 3
D) u21=32/3u _ { 21 } = 32 / 3
E) u21=13/3u _ { 21 } = 13 / 3
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42
In the previous problem, using the notation uij=u(x,t)u _ { i j } = u ( x , t ) , and letting c=1,λ=ck/h2c = 1 , \lambda = c k / h ^ { 2 } , the equation becomes

A) ui,j1=λui+1,j+(1+2λ)ui,j+λui1,ju _ { i , j - 1 } = \lambda u _ { i + 1 , j } + ( 1 + 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
B) ui,j1=λui+1,j+(12λ)ui,j+λui1,ju _ { i , j - 1 } = \lambda u _ { i + 1 , j } + ( 1 - 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
C) ui,j+1=λui+1,j+(1+2λ)ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 + 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
D) ui,j+1=λui+1,j+(12λ)ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
E) ui,j+1=λui+1,j+(1λ)ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
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43
Is the value of λ\lambda in the previous problem such that the scheme is stable?

A) yes
B) no
C) It is right on the borderline.
D) It cannot be determined from the available data.
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44
The Fourier series of an even function can contain Select all that apply.

A) sine terms
B) cosine terms
C) a constant term
D) more than one of the above
E) none of the above
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45
The eigenvalue-eigenvector pairs for the matrix A=(400031011)XA = \left( \begin{array} { c c c } 4 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X } are Select all that apply.

A) 4,(100)4 , \left( \begin{array} { l } 1 \\0 \\0\end{array} \right)
B) 2,(011)2 , \left( \begin{array} { c } 0 \\1 \\- 1\end{array} \right)
C) 2,(110)2 , \left( \begin{array} { c } 1 \\- 1 \\0\end{array} \right)
D) 2,(011)2 , \left( \begin{array} { l } 0 \\1 \\1\end{array} \right)
E) 2,(010)2 , \left( \begin{array} { l } 0 \\1 \\0\end{array} \right)
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46
In the previous two problem, the solution for u(x,t)u ( x , t ) is

A) u=2u00[(1ekα2t)sin(αx)/α]dα/πu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha / \pi
B) u=2u00[(1ekα2t)sin(αx)/α]dαu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha
C) u=u00[(1ekα2t)sin(αx)/α]dαu = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha
D) u=u00[(1ekα2t)sin(αx)/α]dα/πu = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \alpha x ) / \alpha \right] d \alpha / \pi
E) u=u00[(1ekα2t)sin(x)/α]dα/(2π)u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) \sin ( \infty x ) / \alpha \right] d \alpha / ( 2 \pi )
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47
The solutions of a regular Sturm-Liouville problem ((ry)+(λp+q)y=0,y(a)=0,y(b)=0)\left( \left( r y ^ { \prime } \right) ^ { \prime } + ( \lambda p + q ) y = 0 , y ( a ) = 0 , y ( b ) = 0 \right) have which of the following properties?

A) There exists an infinite number of real eigenvalues.
B) The eigenvalues are orthogonal on [a,b][ a , b ] .
C) For each eigenvalue, there is only one eigenfunction (except for non-zero constant multiples).
D) Eigenfunctions corresponding to different eigenvalues are linearly independent.
E) The set of eigenfunctions corresponding to the set of eigenvalues is orthogonal with respect to the weight function r(x)r ( x ) on the interval [a,b][ a , b ] .
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48
Consider the heat problem c2ux2=ut,u(0,t)=0,u(1,t)=3,u(x,0)=3x2c \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , u ( 1 , t ) = 3 , u ( x , 0 ) = 3 x ^ { 2 } . Replace 2ux2\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } with a central difference approximation with h=1/3h = 1 / 3 and ut\frac { \partial u } { \partial t } with a forward difference approximation with k=1/2k = 1 / 2 . The resulting equation is

A) c[u(x+h,t)+2u(x,t)+u(xh,t)]/h2=(u(x,t+k)u(x,t))/kc [ u ( x + h , t ) + 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k
B) c[u(x+h,t)+2u(x,t)+u(xh,t)]/h2=(u(x,t+k)+u(x,t))/kc [ u ( x + h , t ) + 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) + u ( x , t ) ) / k
C) c[u(x+h,t)2u(x,t)+u(xh,t)]/h2=(u(x,t+k)u(x,t))/kc [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k
D) c[u(x+h,t)4u(x,t)+u(xh,t)]/h2=(u(x,t+k)+u(x,t))/kc [ u ( x + h , t ) - 4 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) + u ( x , t ) ) / k
E) c[u(x+h,t)4u(x,t)+u(xh,t)]/h2=(u(x,t+k)u(x,t))/kc [ u ( x + h , t ) - 4 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k
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