The solution of the system of differential equations $\begin{array} { l }
\frac { d x } { d t } = - 6 x + 5 y + t \
\frac { d y } { d t } = - 5 x + 4 y + 1
\end{array}$ is
A) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t - 16$
B) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } + 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t + 16$
C) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { t } + c _ { 2 } t e ^ { t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { t } + c _ { 2 } t e ^ { t } - 5 t + 16$
D) $x = \left( c _ { 1 } - c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t + 16$
E) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t + 16$

Question

The solution of the system of differential equations $\begin{array} { l } 
\frac { d x } { d t } = - 6 x + 5 y + t \
\frac { d y } { d t } = - 5 x + 4 y + 1
\end{array}$ is
A) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t - 16$
B) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } + 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t + 16$
C) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { t } + c _ { 2 } t e ^ { t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { t } + c _ { 2 } t e ^ { t } - 5 t + 16$
D) $x = \left( c _ { 1 } - c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t + 16$
E) $x = \left( c _ { 1 } + c _ { 2 } / 5 \right) e ^ { - t } + c _ { 2 } t e ^ { - t } - 4 t + 69 / 5 , y = c _ { 1 } e ^ { - t } + c _ { 2 } t e ^ { - t } - 5 t + 16$

Quizplus · Accepted Answer

The given system of differential equations can be written in matrix form as $\frac{d}{dt}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} -6 & 5 \\ -5 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}+\begin{bmatrix} t \\ 1 \end{bmatrix}$The eigenvalues of the coefficient matrix are $\lambda_1=-1$ and $\lambda_2=5$. For $\lambda_1=-1$, the corresponding eigenvector is $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. For $\lambda_2=5$, the corresponding eigenvector is $\begin{bmatrix} 5 \\ -1 \end{bmatrix}$. Therefore, the general solution to the system of differential equations is $\begin{bmatrix} x \\ y \end{bmatrix}=c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix}e^{-t}+c_2\begin{bmatrix} 5 \\ -1 \end{bmatrix}e^{5t}+\begin{bmatrix} -\frac{4}{5}t+\frac{69}{5} \\ -5t-16 \end{bmatrix}$where $c_1$ and $c_2$ are constants.

The Solution of the System of Differential Equations dxdt=−6x+5y+tdydt=−5x+4y+1\begin{array} { l } \frac { d x } { d t } = - 6 x + 5 y + t \\\frac { d y } { d t } = - 5 x + 4 y + 1\end{array}dtdx​=−6x+5y+tdtdy​=−5x+4y+1​

The Solution of the System of Differential Equations $\begin{array} { l } \frac { d x } { d t } = - 6 x + 5 y + t \\\frac { d y } { d t } = - 5 x + 4 y + 1\end{array}$