Exam 16: Mathematics Problems: Differential Equations and Linear Algebra

Using Laplace transform methods, the solution of $y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1$ is (Hint: the previous problem might be useful.)

The correct form of the particular solution of $y ^ { \prime \prime } - 2 y ^ { \prime } = x + e ^ { x }$ is

The eigenvalues of the matrix $A = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right)$ are

In the previous problem, the solution of the differential equation is

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

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

The solution of $y ^ { \prime } - y = x$ is

The correct form of the particular solution of $y ^ { \prime \prime } - 2 y ^ { \prime } + y = e ^ { x }$ is

Using the convolution theorem, we find that $\mathcal { L } ^ { - 1 } \left\{ 1 / \left( ( s + 1 ) \left( s ^ { 2 } + 1 \right) \right) \right\} =$

In the previous problem, the solution for the position, $x ( t )$ , is

Consider the boundary-value problem $y ^ { \prime \prime } - 4 y ^ { \prime } + 3 y = x , y ( 0 ) = 1 , y ( 1 ) = 2$ . Replace the derivatives with central differences with a step size of $h = 1 / 4$ . The resulting equations are

The eigenvalues of the matrix $A = \left( \begin{array} { c c c } 3 & 0 & 0 \\0 & 3 & 1 \\0 & - 1 & 1\end{array} \right)$ are

The solution of $x y ^ { \prime } = ( x + 1 ) y ^ { 2 }$

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

The eigenvalues of the matrix $A = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right)$ are

The solution of $y ^ { \prime \prime } + 3 y ^ { \prime } - 4 y = \cos x$ is

The solution of $y ^ { \prime \prime } + 4 y ^ { \prime } + 4 y = 0$ is

A frozen chicken at $0 ^ { \circ } \mathrm { C }$ is taken out of the freezer and placed on a table at $20 ^ { \circ } \mathrm { C }$ . One hour later the temperature of the chicken is $18 ^ { \circ } \mathrm { C }$ . The mathematical model for the temperature $T ( t )$ as a function of time $t$ is (assuming Newton's law of warming)

The solution of $\mathbf { X } ^ { \prime } = \left( \begin{array} { c c } 1 & - 2 \\1 & 4\end{array} \right) \mathbf { X }$ is

The solution of $y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0$ is