Exam 16: Mathematics Problems: Differential Equations and Linear Algebra

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The solution of x2y′′+3xy′−3y=0x ^ { 2 } y ^ { \prime \prime } + 3 x y ^ { \prime } - 3 y = 0 is

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The solution of X′=(3000310−11)XX ^ { \prime } = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right) \mathrm { X } is

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The differential equation y′+y=x2y ^ { \prime } + y = x ^ { 2 } is Select all that apply.

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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)=0y ^ { \prime } = 1 - y ^ { 2 } , y ( 0 ) = 0 at x=0.1x = 0.1 is

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The solution of X′=(1−21−1)X\mathbf { X } ^ { \prime } = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) \mathbf { X } is

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

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Using Laplace transform methods, the solution of y′′+2y′+y=e−t,y(0)=1,y′(0)=0y ^ { \prime \prime } + 2 y ^ { \prime } + y = e ^ { - t } , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is

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In the previous two problems, the solution for the temperature is

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The solution of the previous problem is

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The auxiliary equation of y′′−5y′+6y=0y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0 is

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The solution of x2y′′+xy′=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } = 0 is

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The solution of the system in the previous problem is

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Using Laplace transform methods, the solution of y′′+y=δ(t−π),y(0)=1,y′(0)=0y ^ { \prime \prime } + y = \delta ( t - \pi ) , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0 is

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The solution of x2ydx+(x3/3+y)dy=0x ^ { 2 } y d x + \left( x ^ { 3 } / 3 + y \right) d y = 0 is

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The eigenvalues of the matrix A=(1−21−1)A = \left( \begin{array} { l l } 1 & - 2 \\1 & - 1\end{array} \right) are

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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

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Let A=(200−3)A = \left( \begin{array} { c c } 2 & 0 \\0 & - 3\end{array} \right) . Then eAt=e ^ { A t } =

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The differential equation y′=x2y2y ^ { \prime } = x ^ { 2 } y ^ { 2 } is Select all that apply.

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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)=0y ^ { \prime } = 1 - y ^ { 2 } , y ( 0 ) = 0 at x=0.1x = 0.1 is

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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

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