Chapter 5.7, Problem 1P

Program Plan Intro

Program Description: Purpose of problem is to find a particular solution of the system $x^{\prime }=x+2y+3\text{ and }{y}^{\prime }=2x+y-2$ by using method of undetermined coefficient.

Explanation of Solution

Given information:

The system of differential equation is $x^{\prime }=x+2y+3\text{ and }{y}^{\prime }=2x+y-2$ .

Explanation:

Consider matrix form of the system of differential equation as,

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

Consider $\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$ as matrix A and non homogeneous term $f=\left[\begin{array}{c}3\\ -2\end{array}\right]$

The non homogeneous term is a constant vector.

Therefore, A constant particular solution is in the form as,

  $x_p\left(t\right)=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$

Substitute $\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$ for x and $\left[\begin{array}{l}0\\ 0\end{array}\right]$ for $x^{\prime }$ in equation (1)

  $\begin{array}{c}\left[\begin{array}{l}0\\ 0\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]+\left[\begin{array}{c}3\\ -2\end{array}\right]\\ \left[\begin{array}{l}0\\ 0\end{array}\right]=\left[\begin{array}{l}{a}_1+2a_2\\ 2a_1+a_2\end{array}\right]+\left[\begin{array}{c}3\\ -2\end{array}\right]\\ \left[\begin{array}{l}0\\ 0\end{array}\right]=\left[\begin{array}{l}{a}_1+2a_2+3\\ 2a_1+a_2-2\end{array}\right]\end{array}$

Equate the coefficient to obtain first equation.

  $\begin{array}{c}{a}_1+2a_2+3=0\\ a_1=-3-2a_2\end{array}$

The second equation is obtained as,

  $2a_1+a_2-2=0$

Substitute $-3-2a_2$ for $a_1$ in above equation.

  $\begin{array}{c}2\left(-3-2a_2\right)+a_2-2=0\\ -6-4a_2+a_2-2=0\\ -3a_2=8\\ a_2=-\frac{8}{3}\end{array}$

The value of $a_1$ is obtained as,

  $\begin{array}{c}{a}_1=-3-2\left(-\frac{8}{3}\right)\\ =-3+\frac{16}{3}\\ =\frac{-9+16}{3}\\ =\frac{7}{3}\end{array}$

The particular solution of the system of differential equation is,

  $x_p=\left[\begin{array}{l}x\left(t\right)\\ y\left(t\right)\end{array}\right]$ ...............(2)

Substitute $\frac{7}{3}$ for $x\left(t\right)$ and $-\frac{8}{3}$ for $y\left(t\right)$ in equation (2)

  $x_p=\left[\begin{array}{c}\frac{7}{3}\\ -\frac{8}{3}\end{array}\right]$

Conclusion:

Thus, the particular solution of the differential system $x^{\prime }=x+2y+3\text{ and }{y}^{\prime }=2x+y-2$ is $x\left(t\right)=\frac{7}{3}\text{ and }y\left(t\right)=-\frac{8}{3}$ .