Question

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

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

Short answer

The general solution of the system is

$X\left(t\right)=c_1\left(\begin{array}{c}-\sin 4t\\ \cos 4t\end{array}\right)e^t+c_2\left(\begin{array}{c}\cos 4t\\ \sin 4t\end{array}\right)e^t+\left(\begin{array}{c}0\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)e^{6t}+\left(\begin{array}{c}\frac{4}{17}\\ \frac{1}{17}\end{array}\right)$

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 have

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

Where

$A=\left(\begin{array}{cc}1& -4\\ 4& 1\end{array}\right)$

Now, finding the characteristic equation of the coefficient matrix,

$$\begin{gathered} det(A-\lambda I)=0\to \left|\begin{array}{cc}1-\lambda & -4\\ 4& 1-\lambda \end{array}\right|=0 \\ (1-\lambda )(1-\lambda )-(-16)=0 \\ 1-\lambda -\lambda +{\lambda }^2+16=0 \\ {\lambda }^2-2\lambda +17=0 \\ (1-\lambda {)}^2=-16 \\ \lambda =1\pm 4i \end{gathered}$$

So, our eigenvalues are${\lambda }_1=1+4i,{\lambda }_2=1-4i$

Determine the eigenvector and corresponding solution vector

For ${\lambda }_1=1+4i$:

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

Apply row operation$R_2+R_1\to R_2$

$=\left(\begin{array}{cc}-4i& 4\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Here we get a single equation,

$4i{k}_1-4k_2=0\to k_1=i{k}_2$

And

$4k_1-4i{k}_2=0\to k_1=i{k}_2$

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

role="math" localid="1668156333743" $$\begin{gathered} K_1=\left(\begin{array}{c}i\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)+i\left(\begin{array}{c}1\\ 0\end{array}\right),X_1=\left(\begin{array}{c}1\\ 1\end{array}\right)e^{4t} \\ where \\ {B}_1=\left(\begin{array}{c}0\\ 1\end{array}\right)and{B}_2=\left(\begin{array}{c}1\\ 0\end{array}\right) \end{gathered}$$

Determine the value of XC.

Also $\lambda =\alpha +\beta i\to {\lambda }_1=1+4i$

Where$\alpha =1and\beta =4$

$$\begin{gathered} {\mathbf{X}}_{\mathbf{1}}=\left[{\mathbf{B}}_1\cos \beta t-{\mathbf{B}}_2\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}0\\ 1\end{array}\right)\cos 4t-\left(\begin{array}{c}1\\ 0\end{array}\right)\sin 4t\right]e^t=\left(\begin{array}{c}-\sin 4t\\ \cos 4t\end{array}\right)e^t \end{gathered}$$

And

$$\begin{gathered} {\mathbf{X}}_2=\left[{\mathbf{B}}_2\cos \beta t+{\mathbf{B}}_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}1\\ 0\end{array}\right)\cos 4t+\left(\begin{array}{c}0\\ 1\end{array}\right)\sin 4t\right]e^t=\left(\begin{array}{c}\cos 4t\\ \sin 4t\end{array}\right)e^t \end{gathered}$$

Therefore,

$$\begin{gathered} {\mathbf{X}}_{\mathbf{c}}=c_1{\mathbf{X}}_{\mathbf{1}}+c_2{\mathbf{X}}_{\mathbf{2}} \\ ={\mathrm{c}}_1\left(\begin{array}{c}-\sin 4\mathrm{t}\\ \cos 4\mathrm{t}\end{array}\right){\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2\left(\begin{array}{c}\cos 4\mathrm{t}\\ \sin 4\mathrm{t}\end{array}\right){\mathrm{e}}^{\mathrm{t}} \end{gathered}$$

Determine the value of a and b

Since , $F\left(t\right)=\left(\begin{array}{c}4t+9e^{6t}\\ -t+e^{6t}\end{array}\right)$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)t+\left(\begin{array}{c}{a}_2\\ b2\end{array}\right)e^{6t}+\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)$$X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)e^{6t}+\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)$

Differentiating,

$X{\text{'}}_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}{a}_2\\ b2\end{array}\right)6e^{6t}$

Substituting this last assumption into the given system yields

$\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}{a}_2\\ b2\end{array}\right)6e^{6t}=\left(\begin{array}{cc}1& -4\\ 4& 1\end{array}\right)\left[\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)e^{6t}+\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)\right]+\left(\begin{array}{c}4t+9e^{6t}\\ -t+e^{6t}\end{array}\right)$

$\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}6a_2\\ 6b_2\end{array}\right)e^{6t}=\left[\left(\begin{array}{c}{a}_1-4b_1\\ 4a_1+b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2-4b_2\\ 4a_2+b_2\end{array}\right)e^{6t}+\left(\begin{array}{c}{a}_3-4b_3\\ 4a_3+b_3\end{array}\right)\right]+\left(\begin{array}{c}4t+9e^{6t}\\ -t+e^{6t}\end{array}\right)$

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

$$\begin{gathered} a_1=a_3-4b_3\to \left(1\right) \\ {b}_1=4a_3+b_3\to \left(2\right) \\ 6a_2=a_2-4b_2+9\to \left(3\right) \\ 6b_2=4a_2+b_2+1\to \left(4\right) \\ {a}_1-4b_1+4=0\to \left(5\right) \\ 4a_1+b_1-1=0\to \left(6\right) \end{gathered}$$

Multiplying equation (5) by -4, the equation becomes

$-4a_1+16b_1-16=0\to \left(7\right)$

$a_3-4b_3=0\to \left(8\right)$

we add equation (7) to (6) and solve for b₁,

$17b_1-17=0\to b_1=1$

substituting this value of b₁ into equation (5) or (6) or (7), whatever you prefer).

$a_1-4\left(1\right)+4=0\to a_1=0$

we can now substitute these values we obtained into equations (1) and (2), equation (1) becomes,

$a_3-4b_3=0\to \left(8\right)$

equation (2) becomes,

$4a_3+b_3=1\to \left(9\right)$

multiplying equation (8) by 4, the equation becomes

$4a_3-16b_3=0\to \left(10\right)$

subtracting equation (10) from (9)

$17b_3=1\to b_3=\frac{1}{17}$

We can substitute this value of into (9) to solve for a₃

$4a_3+\frac{1}{17}=1\to a_3=\frac{4}{17}$

From equation (3) we have

$5a_2=-4b_2+9\to a_3=-\frac{4}{5}{b}_2+\frac{9}{5}$

we substitute this into equation (4) and we solve for b₂,

$$\begin{gathered} 6b_2=4\left(-\frac{4}{5}{b}_2+\frac{9}{5}\right)+b_2+1\to \frac{41}{5}{b}_2=\frac{41}{5} \\ {b}_2=1 \end{gathered}$$

so now, to find the value of a₃, we have

$a_2=-\frac{4}{5}\left(1\right)+\frac{9}{5}\to a_2=1$

Determine the general solution of the system

we plug these values into our particular solution, where

$X_p=\left(\begin{array}{c}0\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)e^{6t}+\left(\begin{array}{c}\frac{4}{17}\\ \frac{1}{17}\end{array}\right)$

and we finally conclude that the general solution of the system is

role="math" localid="1668176111142" $$\begin{gathered} \mathbf{X}\left(t\right)={\mathbf{X}}_{\mathbf{c}}+{\mathbf{X}}_{\mathbf{p}} \\ X\left(t\right)=c_1\left(\begin{array}{c}-\sin 4t\\ \cos 4t\end{array}\right)e^t+c_2\left(\begin{array}{c}\cos 4t\\ \sin 4t\end{array}\right)e^t+\left(\begin{array}{c}0\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)e^{6t}+\left(\begin{array}{c}\frac{4}{17}\\ \frac{1}{17}\end{array}\right) \end{gathered}$$