# Find the solution to the given differential equation

I am trying to find the general solution to the system $$x' = Ax + at$$ where,

$$A = \begin{pmatrix} 1 & -1\\ 9 & 1 \end{pmatrix}$$

and $$a = \begin{pmatrix} -1\\ 1 \end{pmatrix}$$

I began by finding the eigenvectors and eigenvalues of $$A$$ and got the eigevalues $$\lambda = 1 \pm 3i$$ to correspond to the eigenvectors $$\begin{pmatrix} i\\ 3 \end{pmatrix}$$ and $$\begin{pmatrix} i\\ -3 \end{pmatrix}$$

I am confused about how to proceed to finding the final solution here. Any guidance is greatly appreciated!

• Hint: variation of constant. Commented May 6, 2021 at 3:08
• Thank you @Moo - I wasn't familiar with applying these methods to matrices. Solved it!
– user815455
Commented May 6, 2021 at 3:29
• Thanks @newbie - your hint helped me solve the problem! I really appreciate it.
– user815455
Commented May 6, 2021 at 3:30

Typically, the general method of solving such problems is the Laplace transform which yields $$x'=Ax+at\implies sx-x(0)=Ax+L(at)\implies (sI-A)x=x(0)+L(at)\implies x=(sI-A)^{-1}(x(0)+L(at)) .$$ In our case, $$x{= \begin{bmatrix} s-1&1\\-9&s-1 \end{bmatrix}^{-1} (x(0)+L(at)) \\ =\begin{bmatrix} \frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9} \end{bmatrix} (x(0)+L(at)) }.$$ If out experiment starts at $$t=0$$, we can substitute $$t$$ with $$r(t)$$ which is the ramp function with the Laplace transform of $$\frac{1}{s^2}$$. Hence $$x =\begin{bmatrix} \frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9} \end{bmatrix} \begin{bmatrix} x_1(0)+\frac{a}{s^2}\\ x_2(0)+\frac{a}{s^2} \end{bmatrix}$$