Chapter 7.9, Problem 1P

To determine

The general solution of the equation, $x^{\prime }=\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)x+\left(\begin{array}{c}{e}^t\\ t\end{array}\right)$.

Answer to Problem 1P

The general solution of $x^{\prime }=\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)x+\left(\begin{array}{c}{e}^t\\ t\end{array}\right)$ is obtained as, $\underset{\_}{x\left(t\right)=c_1\left(\begin{array}{c}1\\ 3\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}1\\ 1\end{array}\right)e^t+\left(\begin{array}{c}\frac{3}{2}t-\frac{1}{4}\\ \frac{3}{2}t-\frac{3}{4}\end{array}\right)e^t+\left(\begin{array}{c}1\\ 2\end{array}\right)t-\left(\begin{array}{c}0\\ 1\end{array}\right)\left(\begin{array}{c}\frac{-2}{3}{e}^t-e^{-t}\\ \frac{-1}{\sqrt{3}}{e}^t+\frac{2}{\sqrt{3}}{e}^{-t}\end{array}\right)}$.

Explanation of Solution

The given system of equations is,

$x^{\prime }=\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)x+\left(\begin{array}{c}{e}^t\\ t\end{array}\right)$

It is noted that, the given system is a non-homogeneous system.

Consider the homogeneous part. That is,

$x^{\prime }=\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)x$

This is of the form, $x^{\prime }=Ax$, where $A=\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)$.

The eigenvalues r and the eigenvectors $\xi$ satisfy the equation, $\left|A-rI\right|\xi =0$. That is,

$\begin{array}{l}\left|A-rI\right|\xi =0\\ \left|\begin{array}{cc}2-r& -1\\ 3& -2-r\end{array}\right|\left(\begin{array}{c}{\xi }_1\\ {\xi }_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(2-r\right)\left(-2-r\right)+3=0\\ r^2-1=0\end{array}$

The roots of the above characteristic equation is obtained as follows.

$\begin{array}{l}{r}^2-1=0\\ r^2=1\end{array}$

Thus, the solutions are, $r_1=-1,r_2=1$. That is, the solutions are real and different.

Set $r_1=-1$ in the coefficient matrix. That is,

$\left(\begin{array}{cc}2+1& -1\\ 3& -2+1\end{array}\right)=\left(\begin{array}{cc}3& -1\\ 3& -1\end{array}\right)$

On solving, $\left(\begin{array}{cc}3& -1\\ 3& -1\end{array}\right)\left(\begin{array}{c}{\xi }_1\\ {\xi }_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$, the following equation is obtained.

$3{\xi }_1-{\xi }_2=0$

Let ${\xi }_1=1$, then the above equation becomes, ${\xi }_2=3$.

Then, the eigen vector formed is, $\xi =\left(\begin{array}{c}1\\ 3\end{array}\right)$.

Thus, the first solution of the homogeneous solution is, ${\xi }^{\left(1\right)}=\left(\begin{array}{c}1\\ 3\end{array}\right)c_2{e}^{-t}$.

Set $r_2=1$ in the coefficient matrix. That is,

$\left(\begin{array}{cc}2-1& -1\\ 3& -2-1\end{array}\right)=\left(\begin{array}{cc}1& -1\\ 3& -3\end{array}\right)$

On solving, $\left(\begin{array}{cc}1& -1\\ 3& -3\end{array}\right)\left(\begin{array}{c}{\xi }_1\\ {\xi }_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$, the following equation is obtained.

$\begin{array}{l}{\xi }_1-{\xi }_2=0\\ {\xi }_1={\xi }_2\end{array}$

Let ${\xi }_1=1$, then from the above equation, ${\xi }_2=1$.

Then, the eigen vector formed is, $\xi =\left(\begin{array}{c}1\\ 1\end{array}\right)$.

Thus, the second solution of the homogeneous solution is, ${\xi }^{\left(2\right)}=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^t$.

Therefore, the matrix T of eigen vectors is,

$\begin{array}{c}T=\left({\xi }^{\left(1\right)},{\xi }^{\left(2\right)}\right)\\ =\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\end{array}$

Then, the inverse of T is, $T^{-1}=-\frac{1}{2}\left(\begin{array}{cc}1& -1\\ -3& 1\end{array}\right)$.

It is noted from the given equation that, $g\left(t\right)=\left(\begin{array}{c}{e}^t\\ t\end{array}\right)$.

Now consider the equation, $x=Ty$. Then, the following equation is obtained.

$y^{\prime }=Dy+T^{-1}g\left(t\right)\dots \left(1\right)\text{ where, }D=\left(\begin{array}{cc}{r}_1& 0\\ 0& r_2\end{array}\right)$

Substitute the eigen values and inverse of T in the equation (1) as follows.

$\begin{array}{l}{y}^{\prime }=Dy+T^{-1}g\left(t\right)\\ \left(\begin{array}{c}{y}_1{}^{\prime }\\ y_2{}^{\prime }\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right)-\frac{1}{2}\left(\begin{array}{cc}1& -1\\ -3& 1\end{array}\right)\left(\begin{array}{c}{e}^t\\ t\end{array}\right)\end{array}$

Thus, the following two linear differential equations are obtained.

$\begin{array}{l}{y}_1{}^{\prime }=-y_1-\frac{1}{2}{e}^t+\frac{1}{2}t\dots \left(2\right)\\ y_2{}^{\prime }=y_2+\frac{3}{2}{e}^t-\frac{1}{2}t\dots \left(3\right)\end{array}$

Consider the equation (2). The integral factor is,

$\begin{array}{c}\mu \left(t\right)=e^{{\displaystyle \int p\left(t\right)dt}}\\ =e^{-t}\end{array}$

Multiply the obtained integral factor throughout the equation (2) and $y_1$ is obtained as,

$y_1=c_1{e}^{-t}-\frac{1}{4}{e}^t+\frac{1}{2}t-\frac{1}{2}\dots \left(4\right)$

Similarly, multiply the obtained integral factor $e^t$ throughout the equation (3) and $y_2$ is obtained as,

$y_2=\frac{3}{2}t{e}^t+\frac{1}{2}t+\frac{1}{2}{\text{+c}}_2{e}^t\dots \left(5\right)$

The solution is given by the equation, $x=Ty$. That is, $x\left(t\right)=T\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right)\dots \left(6\right)$.

Substitute the obtained solutions in the equation (6) as follows.

$\begin{array}{c}x\left(t\right)=\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right)\\ =\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\left(\begin{array}{c}{c}_1{e}^{-t}-\frac{1}{4}{e}^t+\frac{1}{2}t-\frac{1}{2}\\ \frac{3}{2}t{e}^t+\frac{1}{2}t+\frac{1}{2}{\text{+c}}_2{e}^t\end{array}\right)\\ =\left(\begin{array}{c}{c}_1{e}^{-t}-\frac{1}{4}{e}^t+\frac{1}{2}t-\frac{1}{2}+\frac{3}{2}t{e}^t+\frac{1}{2}t+\frac{1}{2}{\text{+c}}_2{e}^t\\ 3\left(c_1{e}^{-t}-\frac{1}{4}{e}^t+\frac{1}{2}t-\frac{1}{2}\right)+\frac{3}{2}t{e}^t+\frac{1}{2}t+\frac{1}{2}{\text{+c}}_2{e}^t\end{array}\right)\\ =c_1\left(\begin{array}{c}1\\ 3\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}1\\ 1\end{array}\right)e^t+\left(\begin{array}{c}\frac{3}{2}t-\frac{1}{4}\\ \frac{3}{2}t-\frac{3}{4}\end{array}\right)e^t+\left(\begin{array}{c}1\\ 2\end{array}\right)t-\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}$

Therefore, the general solution of $x^{\prime }=\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)x+\left(\begin{array}{c}{e}^t\\ t\end{array}\right)$ is obtained as, $\underset{\_}{x\left(t\right)=c_1\left(\begin{array}{c}1\\ 3\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}1\\ 1\end{array}\right)e^t+\left(\begin{array}{c}\frac{3}{2}t-\frac{1}{4}\\ \frac{3}{2}t-\frac{3}{4}\end{array}\right)e^t+\left(\begin{array}{c}1\\ 2\end{array}\right)t-\left(\begin{array}{c}0\\ 1\end{array}\right)\left(\begin{array}{c}\frac{-2}{3}{e}^t-e^{-t}\\ \frac{-1}{\sqrt{3}}{e}^t+\frac{2}{\sqrt{3}}{e}^{-t}\end{array}\right)}$.