Question

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

$X\text{'}=\left(\begin{array}{cc}1& 3\\ 3& 1\end{array}\right)X+\left(\begin{array}{c}-2t^2\\ t+5\end{array}\right)$

Short answer

The general solution of the system is

$X\left(t\right)=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{4t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-2t}+\left(\begin{array}{c}-\frac{1}{4}\\ \frac{3}{4}\end{array}\right)t^2+\left(\begin{array}{c}\frac{1}{4}\\ -\frac{1}{4}\end{array}\right)t+\left(\begin{array}{c}-2\\ \frac{3}{4}\end{array}\right)$

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

$X\text{'}=\left(\begin{array}{cc}1& 3\\ 3& 1\end{array}\right)X+\left(\begin{array}{c}-2t^2\\ t+5\end{array}\right)$

Where

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

Now, finding the characteristic equation of the coefficient matrix,

$$\begin{gathered} det(A-\lambda I)=0\to \left|\begin{array}{cc}1-\lambda & 3\\ 3& 1-\lambda \end{array}\right|=0 \\ (1-\lambda )(1-\lambda )-9=0 \\ 1-\lambda -\lambda +{\lambda }^2-9=0 \\ {\lambda }^2-2\lambda +8=0 \\ (\lambda -4)(\lambda +2)=0 \end{gathered}$$

So, our eigenvalues are${\lambda }_1=4,{\lambda }_2=-2$

Determine the eigenvector and corresponding solution vector

For${\lambda }_1=4$:

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

Apply row operation$R_2-R_1\to R_2$

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

Here we get a single equation,

$$\begin{gathered} -3k_1+3k_2=0\to -3k_1=-3k_2 \\ {k}_1=k_2 \end{gathered}$$

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

$K_1=\left(\begin{array}{c}1\\ 1\end{array}\right),X_1=\left(\begin{array}{c}1\\ 1\end{array}\right)e^{4t}$

Determine the eigenvector and corresponding solution vector

For${\lambda }_2=-2$:

$$\begin{gathered} (A+2I)K=0\to \left(\begin{array}{cc}1-(-2)& 3\\ 3& 1-(-2)\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ =\left(\begin{array}{cc}3& 3\\ 3& 3\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation$R_2-R_1\to R_2$

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

Here we get a single equation,

$$\begin{gathered} 3k_1+3k_2=0\to 3k_1=-3k_2 \\ {k}_1=-k_2 \end{gathered}$$

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

$K_2=\left(\begin{array}{c}1\\ -1\end{array}\right),X_2=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-2t}$

Determine the general solution of the system

Therefore,

$X=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{4t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-2t}$

Since$F\left(t\right)=\left(\begin{array}{c}-2t^2\\ t+5\end{array}\right)$, we shall try to find a particular solution of the system that possesses the same form:

$X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)t^2+\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)$

Differentiating,

$X{\text{'}}_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)2t$

Substituting this last assumption into the given system yields

$$\begin{gathered} \left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)2t=\left(\begin{array}{cc}1& 3\\ 3& 1\end{array}\right)\left[\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)t^2+\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)\right]+\left(\begin{array}{c}-2t^2\\ t+5\end{array}\right) \\ \left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)+\left(\begin{array}{c}2a_2\\ 2b_2\end{array}\right)t=\left(\begin{array}{c}{a}_1+3b_1\\ 3a_1+b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2+3b_2\\ 3a_2+b_2\end{array}\right)t^2+\left(\begin{array}{c}{a}_3+3b_3\\ 3a_3+b_3\end{array}\right)+\left(\begin{array}{c}-2t^2\\ t+5\end{array}\right) \end{gathered}$$

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

$$\begin{gathered} a_1=a_3+3b_3\to \left(1\right) \\ {b}_1=3a_3+b_3+5\to \left(2\right) \\ 2a_2=a_1+3b_1\to \left(3\right) \\ 2b_2=3a_1+b_1+1\to \left(4\right) \\ {a}_2+3b_2-2=0\to \left(5\right) \\ 3a_2+b_2=0\to \left(6\right) \end{gathered}$$

From the (6) we have

$3a_2+b_2=0\to b_2=-3a_2$

we can substitute this into (5), where

$$\begin{gathered} a_2+3\left(-3a_2\right)-2=0 \\ {a}_2-9a_2-2=0 \\ -8a_2-2=0 \\ {a}_2=-\frac{1}{4} \end{gathered}$$

Using the value we obtained for a₂ to get the value of b₂

_{$b_2=-3\left(-\frac{1}{4}\right)=\frac{3}{4}$}

Now that we have the values of a₂ and b₂, substituting these values into (4)

And (3)

equation (3) becomes:

$-\frac{1}{2}=a_1+3b_1\to \left(7\right)$

Equation (4) becomes:

$\frac{1}{2}=3a_1+b_1\to \left(8\right)$

We can multiply (7)by 3

$-\frac{3}{2}=3a_1+9b_1\to \left(9\right)$

Determine the general solution of the system

now subtracting (9) from (8) to solve for b₁

$2=-8b_1\to b_1=-\frac{1}{4}$

now that we have the value of b₁, we can substitute it into equation(7) to obtain the value of a₁

$-\frac{1}{2}=a_1+3\left(-\frac{1}{4}\right)\to a_1=\frac{1}{4}$

finally, we substitute the values of a₁and b₁ into equations (!) and (2), which will yields new equations,

$$\begin{gathered} \frac{1}{4}=a_3+3b_3\to \left(10\right) \\ -\frac{21}{4}=3a_3+b_3\to \left(11\right) \end{gathered}$$

Multiplying equation (10) by 3yields

$\frac{3}{4}=3a_3+9b_3\to \left(12\right)$

Subtracting equation (11) from (12) to solve

$6=8b_3\to b_3=\frac{3}{4}$

we can substitute this value into equation(10) to obtain the value of a₃

$\frac{1}{4}=a_3+3\left(\frac{3}{4}\right)\to a_3=-2$

We plug these values into our particular, where

$$\begin{gathered} X_p=\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)t+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)t^2+\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right) \\ =\left(\begin{array}{c}-\frac{1}{4}\\ \frac{3}{4}\end{array}\right)t^2+\left(\begin{array}{c}\frac{1}{4}\\ -\frac{1}{4}\end{array}\right)t+\left(\begin{array}{c}-2\\ \frac{3}{4}\end{array}\right) \end{gathered}$$

We finally conclude that the general solution of the system is

$X\left(t\right)=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{4t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-2t}+\left(\begin{array}{c}-\frac{1}{4}\\ \frac{3}{4}\end{array}\right)t^2+\left(\begin{array}{c}\frac{1}{4}\\ -\frac{1}{4}\end{array}\right)t+\left(\begin{array}{c}-2\\ \frac{3}{4}\end{array}\right)$