Question

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

$$\begin{gathered} 2.\frac{dx}{dt}=5x+9y+2 \\ \frac{dy}{dt}=-x+11y+6 \end{gathered}$$

Short answer

The solution of nonhomogeneous system $\left\{\begin{array}{l}\frac{dx}{dt}=5x+9y+2\\ \frac{dy}{dt}=-x+11y+6\end{array}\right.$ is$x\left(t\right)=c_1\left(\begin{array}{c}3\\ 1\end{array}\right)e^{8t}+c_2\left[\left(\begin{array}{c}3\\ 1\end{array}\right)t{e}^{8t}+\left(\begin{array}{c}2\\ 1\end{array}\right)e^{8t}\right]+\left(\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)$

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

$\left\{\begin{array}{l}\frac{dx}{dt}=5x+9y+2\\ \frac{dy}{dt}=-x+11y+6\end{array}\right.$

Which can be written in the form

$$\begin{gathered} X^{\text{'}}=\left(\begin{array}{cc}5& 9\\ -1& 11\end{array}\right)X+\left(\begin{array}{c}2\\ 6\end{array}\right) \\ A=\left(\begin{array}{cc}5& 9\\ -1& 11\end{array}\right) \end{gathered}$$

Now, finding the characteristic equation of the coefficient matrix,

$$\begin{gathered} det(A-\lambda I)=0\to \left|\begin{array}{cc}5-\lambda & 9\\ -1& 11-\lambda \end{array}\right|=0 \\ (5-\lambda )(11-\lambda )-(-9)=0 \\ 55-5\lambda -11\lambda +{\lambda }^2+9=0 \\ {\lambda }^2-16\lambda +64=0 \\ (\lambda -8)(\lambda -8)=0 \end{gathered}$$

So, our eigenvalues are $\lambda =8$

Determine the eigenvector and corresponding solution vector

For $\lambda =8$:

$$\begin{gathered} (A-8I)K=0\to \left(\begin{array}{cc}5-8& 9\\ -1& 11-8\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ =\left(\begin{array}{cc}-3& 9\\ -1& 3\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

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

$=\left(\begin{array}{cc}-3& 9\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Here we get a single equation,

$-3k_1+9k_2=0\to k_1=3k_2$

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

$K=\left(\begin{array}{c}3\\ 1\end{array}\right),X_1=\left(\begin{array}{c}3\\ 1\end{array}\right)e^{8t}$

Determine the eigenvector and corresponding solution vector

For $\lambda =8$:

$$\begin{gathered} \left(A-8\right)K=0\to \left(\begin{array}{cc}5-8& 9\\ -1& 11-8\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}3\\ 1\end{array}\right) \\ =\left(\begin{array}{cc}-3& 9\\ -1& 3\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}3\\ 1\end{array}\right) \end{gathered}$$

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

$=\left(\begin{array}{cc}-3& 9\\ 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}3\\ 0\end{array}\right)$

Here we get a single equation,

$-3k_1+9k_2=3\to k_1=3k_2-1$

Choosing k₂= 1 yields k₁= 2. This gives an eigenvector and a corresponding solution vector:

$P=\left(\begin{array}{c}2\\ 1\end{array}\right),X_2=\left(\begin{array}{c}3\\ 1\end{array}\right)t{e}^{8t}+\left(\begin{array}{c}2\\ 1\end{array}\right)e^{8t}$

Determine the general solution of the system

Therefore,

$X_c=c_1\left(\begin{array}{c}3\\ 1\end{array}\right)e^{8t}+c_2\left[\left(\begin{array}{c}3\\ 1\end{array}\right)t{e}^{8t}+\left(\begin{array}{c}2\\ 1\end{array}\right)e^{8t}\right]$

Since $F\left(t\right)=\left(\begin{array}{c}2\\ 6\end{array}\right)$, 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,

$X_p^{\text{'}}=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Substituting this last assumption into the given system yields

$$\begin{gathered} \left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{cc}5& 9\\ -1& 11\end{array}\right)\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}2\\ 6\end{array}\right) \\ =\left(\begin{array}{c}5a_1+9b_1\\ -a_1+11b_1\end{array}\right)+\left(\begin{array}{c}2\\ 6\end{array}\right) \end{gathered}$$

Where

$$\begin{gathered} 5a_1+9b_1=-2\to \left(1\right) \\ -a_1+11b_1=-6\to \left(2\right) \\ -5a_1-55b_1=-30 \end{gathered}$$

Adding the equation we multiplied to equation (1),

$$\begin{gathered} 55b_1+9b_1=-30-2\to 64b_1=-32 \\ {b}_1=-\frac{1}{2} \end{gathered}$$

Substituting the value of b₁into equation (1) or (2) to obtain the value of a₁

$$\begin{gathered} 5a_1+9(-\frac{1}{2})=-2\to 5a_1=\frac{5}{2} \\ {a}_1=\frac{1}{2} \end{gathered}$$

The particular solution is then

$X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)=\left(\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)$

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

$$\begin{gathered} X\left(t\right)=X_c+X_p \\ X\left(t\right)=c_1\left(\begin{array}{c}3\\ 1\end{array}\right)e^{8t}+c_2\left[\left(\begin{array}{c}3\\ 1\end{array}\right)t{e}^{8t}+\left(\begin{array}{c}2\\ 1\end{array}\right)e^{8t}\right]+\left(\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right) \end{gathered}$$