# Particular solution of system of nonhomogeneus differential equations

I am asked to find the solution of the following system of nonhomogeneus differential equations using elimination method:

$$\begin{cases} x'(t)=4x(t)-2y(t)+t^{-3} \\ y'(t)=8x(t)-4y(t)-t^{-2} \end{cases}$$

By differentiating the first equation and substituing in values from second one, I find the characteristic value which is $$\lambda_{1,2} = 0$$, but from there on I am not sure how to proceed.

I believe the solution of the homogeneus equation for $$x(t)$$ is $$x_{0}(t) = C_1e^{0t} = C_1$$, but I do not know in what form should I seek a particular solution. Some help or hints on what to do would be greatly appreciated. Thank you.

• There is a mistake in the system an x instead of an y May 15, 2020 at 18:59
• Oh yes, thank you, I corrected it now. May 15, 2020 at 19:04

$$\begin{cases} x'(t)=4x(t)-2y(t)+t^{-3} \\ y'(t)=8x(t)-4y(t)-t^{-2} \end{cases}$$ Substract both DEs $$2x'-y'=2t^{-3}+t^{-2}$$ Integrate: $$2x-y=-t^{-2}-t^{-1}+C_1$$ $$4x-2y=-2t^{-2}-2t^{-1}+2C_1$$ Now eliminate $$4x-2y$$ from first equation and solve for $$x(t)$$. It's a first DE.
If you have 2 times an eignanvalue equals zero then the solution should be $$c_1+c_2t$$ But you didnt post your attempt so we can check it.