Question

In problem use the method of undetermined coefficients to solve the given nonhomogeneous system.

$$\begin{gathered} 1.\frac{dx}{dt}=2x+3y-7 \\ \frac{dy}{dt}=-x-2y+5 \end{gathered}$$

Short answer

The nonhomogeneous system for $X\left(t\right)=c_1\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t+c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)e^{-t}+\left(\begin{array}{c}-1\\ 3\end{array}\right)$is $\left\{\begin{array}{l}\frac{dx}{dt}=2x+3y-7\\ \frac{dy}{dt}=-x-2y+5\end{array}\right.$

Step-by-step solution

 The Method of undetermined coefficients

The technique of indeterminate coefficients is a method for finding a specific solution to nonhomogeneous ordinary differential equations and recurrence relations in mathematics.

the general solution of the system is

$\mathbf{X}\left(t\right)={\mathbf{X}}_{\mathbf{c}}+{\mathbf{X}}_{\mathbf{p}}$

Determine the eigenvalues:

We are given

$\left\{\begin{array}{l}\frac{dx}{dt}=2x+3y-7\\ \frac{dy}{dt}=-x-2y+5\end{array}\right.$

Which can be written in the form

$$\begin{gathered} X\text{'}=\left(\begin{array}{cc}2& 3\\ -1& -2\end{array}\right)X+\left(\begin{array}{c}-7\\ 5\end{array}\right) \\ A=\left(\begin{array}{cc}2& 3\\ -1& -2\end{array}\right) \end{gathered}$$

Now, finding the characteristic equation of the coefficient matrix,

$$\begin{gathered} det\left(A-\lambda I\right)=0\to \left|\begin{array}{cc}2-\lambda & 3\\ -1& -2-\lambda \end{array}\right| \\ (2-\lambda )(-2-\lambda )-(-3)=0 \\ -4-2\lambda +2\lambda +{\lambda }^2+3=0 \\ {\lambda }^2-1=0 \\ (\lambda -1)(\lambda +1)=0 \end{gathered}$$

So, our eigenvalues are localid="1668146108171" ${\lambda }_1=1and{\lambda }_2=-1$

Determine the eigenvector and corresponding solution vector

For ${\lambda }_1=1$:

$$\begin{gathered} \left(A-I|0\right)=\left(\begin{array}{ccc}2-1& 3& 0\\ -1& -2-1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1& 3& 0\\ -1& -3& 0\end{array}\right) \end{gathered}$$

Apply row operation $R_2+R_1\to R_2$

$=\left(\begin{array}{ccc}1& 3& 0\\ 0& 0& 0\end{array}\right)$

Here we get a single equation,

$k_1+3k_2=0\to k_1=-3k_2$

Choosing k₂=1 yields k₁=-3 . This gives an eigenvector and a corresponding solution vector:

$K_1=\left(\begin{array}{c}-3\\ 1\end{array}\right),X_1=\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t$

Determine the eigenvector and corresponding solution vector

For ${\lambda }_2=-1$:

$$\begin{gathered} \left(A-I|0\right)=\left(\begin{array}{ccc}2-(-1)& 3& 0\\ -1& -2-(-1)& 0\end{array}\right) \\ =\left(\begin{array}{ccc}3& 3& 0\\ -1& -1& 0\end{array}\right) \end{gathered}$$

Apply row operation 3$R_2+R_1\to R_2$

$=\left(\begin{array}{ccc}3& 3& 0\\ 0& 0& 0\end{array}\right)$

Here we get a single equation,

$3k_1+3k_2=0\to k_1=-k_2$

Choosing k₂ =1 yields k₁ =-1. This gives an eigenvector and a corresponding solution vector:

$K_2=\left(\begin{array}{c}-1\\ 1\end{array}\right);X_1=\left(\begin{array}{c}-1\\ 1\end{array}\right)e^{-t}$

Therefore,

$X_2=c_1\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t+c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)e^{-t}$

Determine the general solution of the system

Since $F\left(t\right)=\left(\begin{array}{c}-7\\ 5\end{array}\right)$, we shall try to find a particular solution of the system that possesses the same form:

$X_P=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)$

Differentiating,

localid="1668146958320" $X_P^{\text{'}}=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Substituting this last assumption into the given system yields

$$\begin{gathered} \left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{cc}2& 3\\ -1& -2\end{array}\right)\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}-7\\ 5\end{array}\right) \\ =\left(\begin{array}{c}2a_1+3b_1\\ -a_1-2b_1\end{array}\right)+\left(\begin{array}{c}-7\\ 5\end{array}\right) \end{gathered}$$

Where

$$\begin{gathered} 2a_1+3b_1=7 \\ -a_1-2b_1=-5 \\ -2a_1-4b_1=-10 \end{gathered}$$

Adding this to equation (1)

$$\begin{gathered} -4b_1+3b_1=-10+7\to -b_1=-3 \\ {b}_1=3 \end{gathered}$$

Substituting the value of into equation (1) or (2) to obtain the value of a₁

$$\begin{gathered} 2a_1+3\left(3\right)=7\to 2a_1=-2 \\ {a}_1=-1 \end{gathered}$$

The particular solution is then

$X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)=\left(\begin{array}{c}-1\\ 3\end{array}\right)$

And finally, we conclude that the general solution of the system is

$$\begin{gathered} \mathbf{X}\left(t\right)={\mathbf{X}}_{\mathbf{c}}+{\mathbf{X}}_{\mathbf{p}} \\ X\left(t\right)=c_1\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t+c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)e^{-t}+\left(\begin{array}{c}-1\\ 3\end{array}\right) \end{gathered}$$