Exam 17: Mathematical Problems and Solutions

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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

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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

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The solution of y+y=tanxy ^ { \prime \prime } + y = \tan x is

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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

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The solutions for λ,R and Θ\lambda , R \text { and } \Theta from the previous problem are

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In the previous problem, the exact solution of the initial value problem is

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The solution of y+2y=x+exy ^ { \prime \prime } + 2 y ^ { \prime } = x + e ^ { x } is

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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.

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The solution of y+y=xy ^ { \prime } + y = x is

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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

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In the previous two problems, the error in the improved Euler method at x=0.1x = 0.1 is

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

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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

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In the previous problem, the solution for the temperature is

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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.

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Using power series methods, the solution of xyxy+y=0x y ^ { \prime \prime } - x y ^ { \prime } + y = 0 is

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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

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The solution of y4y+20y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 20 y = 0 is

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The solution of x2yxy=0x ^ { 2 } y ^ { \prime \prime } - x y ^ { \prime } = 0 is

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The solutions of the eigenvalue problem and the other problem from the previous problem are

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