Find a particular solution of a system of nonhomogeneous differential equations

Question

I'm trying to better understand using the method of underdetermined coefficients to solve a system of DEs. Here's an example of one I'm stuck on at the moment:

Find a particular solution $x_p(t)$ of this nonhomogeneous systems: $$x' = \begin{pmatrix}1&0\\ 2&-3\end{pmatrix} x + \begin{pmatrix} 5e^{3t} \\ 3\end{pmatrix}$$

I think I've arrived at an answer, so hopefully someone can verify if I'm correct:

$$x_p(t) = \begin{pmatrix}5 \\ 2\end{pmatrix} e^{2t} + \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

Is this right?

Answer 1

We are given:

$$x' = \begin{bmatrix}1 & 0\\ 2 & -3 \end{bmatrix}x + \begin{bmatrix}5e^{3t}\\ 3\end{bmatrix}$$

We find:

$$x_h(t) = c_1 e^{-3t} \begin{bmatrix}0\\1\end{bmatrix} + c_2e^t\begin{bmatrix}1\\ \dfrac{1}{2}\end{bmatrix}$$

This gives us:

$$\phi(t) = \left( \begin{array}{cc} 0 & e^t \\ e^{-3 t} & \frac{e^t}{2} \\ \end{array} \right)$$

$$\phi^{-1}(t) = \left( \begin{array}{cc} -\frac{e^{3 t}}{2} & e^{3 t} \\ \frac{e^{-t}}{2} & 0 \\ \end{array} \right)$$

So,

$$\phi^{-1}(t).\left( \begin{array}{c} 5 \exp (3 t) \\ 3 \\ \end{array} \right) = \left( \begin{array}{c} 3 e^{3 t}-\frac{5 e^{6 t}}{2} \\ \frac{5 e^{2 t}}{2} \\ \end{array} \right)$$

Integrating this previous result yields:

$$\left( \begin{array}{c} e^{3 t}-\frac{5 e^{6 t}}{12} \\ \frac{5 e^{2 t}}{4} \\ \end{array} \right)$$

This yields:

$$x_p(t) = \phi(t).\left( \begin{array}{c} e^{3 t}-\frac{5 e^{6 t}}{12} \\ \frac{5 e^{2 t}}{4} \\ \end{array} \right) = \left( \begin{array}{c} \frac{5 e^{3 t}}{2} \\ 1+\frac{5 e^{3 t}}{6} \\ \end{array} \right)$$

The final result is:

$$x(t) = x_h(t) + x_p(t)$$

This yields:

$$x(t) = c_1 e^t+\frac{5 e^{3 t}}{2} \\ y(t) = \dfrac{1}{2} c_1 \left(e^{4 t}-1\right) e^{-3 t}+c_2 e^{-3 t}+\dfrac{5 e^{3 t}}{6}+1$$

Answer 2

You answer is incorrect.

How can you get a particular solution with $e^{2t}$ if your problem has $e^{3t}$?

A right way to do it is to represent $$ \begin{bmatrix} 5e^{3t}\\ 3 \end{bmatrix}= \begin{bmatrix} 5\\ 0 \end{bmatrix}e^{3t}+\begin{bmatrix} 0\\ 3 \end{bmatrix}=f_1(t)+f_2(t) $$ and first look for a solution of $$ \dot x=Ax+f_1(t) $$ in the form $$ \begin{bmatrix} a\\ b \end{bmatrix}e^{3t}, $$ and then for a solution of $$ \dot x=Ax+f_2(t) $$in the form $$ \begin{bmatrix} a\\ b \end{bmatrix}. $$ Both problems are elementary and can be solved for a few minutes.