Question

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

$X\text{'}=\left(\begin{array}{cc}4& \frac{1}{3}\\ 9& 6\end{array}\right)X+\left(\begin{array}{c}-3\\ 10\end{array}\right)e^t$

Short answer

The general solution of the system is

$X\left(t\right)=c_1\left(\begin{array}{c}1\\ 9\end{array}\right)e^{7t}+c_2\left(\begin{array}{c}1\\ -3\end{array}\right)e^{3t}+\left(\begin{array}{c}\frac{55}{36}\\ -\frac{19}{4}\end{array}\right)e^t$

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 have

$X\text{'}=\left(\begin{array}{cc}4& \frac{1}{3}\\ 9& 6\end{array}\right)X+\left(\begin{array}{c}-3\\ 10\end{array}\right)e^t$

Where

$A=\left(\begin{array}{cc}4& \frac{1}{3}\\ 9& 6\end{array}\right)$

Now, finding the characteristic equation of the coefficient matrix,

$$\begin{gathered} det(A-\lambda I)=0\to \left|\begin{array}{cc}4-\lambda & \frac{1}{3}\\ 9& 6-\lambda \end{array}\right|=0 \\ (4-\lambda )(6-\lambda )-\frac{9}{3}=0 \\ 24-4\lambda -6\lambda +{\lambda }^2-3=0 \\ {\lambda }^2-10\lambda +21=0 \\ (\lambda -7)(\lambda -3)=0 \end{gathered}$$

So, our eigenvalues are${\lambda }_1=7,{\lambda }_2=3$

Determine the eigenvector and corresponding solution vector

For${\lambda }_1=7$:

$$\begin{gathered} (A-7I)K=0\to \left(\begin{array}{cc}4-7& \frac{1}{3}\\ 9& 6-7\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ =\left(\begin{array}{cc}-3& \frac{1}{3}\\ 9& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

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

$=\left(\begin{array}{cc}-3& \frac{1}{3}\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Here we get a single equation,

$$\begin{gathered} -3k_1+\frac{1}{3}{k}_2=0\to -3k_1=-\frac{1}{3}{k}_2 \\ {k}_1=\frac{1}{9}{k}_2 \end{gathered}$$

Choosing k₂=9 yields k₁= 1.This gives an eigenvector and a corresponding solution vector:

$K_1=\left(\begin{array}{c}1\\ 9\end{array}\right),X_1=\left(\begin{array}{c}1\\ 9\end{array}\right)e^{7t}$

Determine the eigenvector and corresponding solution vector

For${\lambda }_2=3$:

$$\begin{gathered} (A-3I)K=0\to \left(\begin{array}{cc}4-3& \frac{1}{3}\\ 9& 6-3\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ =\left(\begin{array}{cc}1& \frac{1}{3}\\ 9& 3\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation$\frac{1}{9}{R}_2-R_1\to R_2$

$=\left(\begin{array}{cc}1& \frac{1}{3}\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Here we get a single equation,

$k_1+\frac{1}{3}{k}_2=0\to k_1=-\frac{1}{3}{k}_2$

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

$K_2=\left(\begin{array}{c}1\\ -3\end{array}\right),X_2=\left(\begin{array}{c}1\\ -3\end{array}\right)e^{3t}$

Determine the general solution of the system

Therefore,

$X=c_1\left(\begin{array}{c}1\\ 9\end{array}\right)e^{7t}+c_2\left(\begin{array}{c}1\\ -3\end{array}\right)e^{3t}$

Since$F\left(t\right)=\left(\begin{array}{c}-3\\ 10\end{array}\right)e^t$, we shall try to find a particular solution of the system that possesses the same form:

$X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^t$

Differentiating,

$X{\text{'}}_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^t$

Substituting this last assumption into the given system yields

$$\begin{gathered} \left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^t=\left(\begin{array}{cc}4& \frac{1}{3}\\ 9& 6\end{array}\right)\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^t+\left(\begin{array}{c}-3\\ 10\end{array}\right)e^t \\ \left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^t=\left(\begin{array}{c}4a_1+\frac{1}{3}{b}_1\\ 9a_1+6b_1\end{array}\right)e^t+\left(\begin{array}{c}-3\\ 10\end{array}\right)e^t \end{gathered}$$

Compare the terms on each side of the equation we obtained above, and equate the similar terms, where we will obtain

$$\begin{gathered} a_1=4a_1+\frac{1}{3}{b}_1-3\to 3a_1+\frac{1}{3}{b}_1=3\to \left(1\right) \\ {b}_1=9a_1+6b_1+10\to 9a_1+5b_1=-10\to \left(2\right) \end{gathered}$$

Multiply equation (1) by 3, the equation becomes

$9a_1+b_1=9\to \left(3\right)$

Subtract equation (3) from (2)

$$\begin{gathered} 5b_1-b_1=-10-9\to 4b_1=-19 \\ {b}_1=-\frac{19}{4} \end{gathered}$$

To solve for a₁we substitute this value into equation (3)

$$\begin{gathered} 9a_1=-\frac{19}{4}=9\to 9a_1=\frac{55}{4} \\ {b}_1=\frac{55}{36} \end{gathered}$$

The particular solution is then

$X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^t=\left(\begin{array}{c}\frac{55}{36}\\ -\frac{19}{4}\end{array}\right)e^t$

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

$X\left(t\right)=c_1\left(\begin{array}{c}1\\ 9\end{array}\right)e^{7t}+c_2\left(\begin{array}{c}1\\ -3\end{array}\right)e^{3t}+\left(\begin{array}{c}\frac{55}{36}\\ -\frac{19}{4}\end{array}\right)e^t$