I got this system $$ \left\{ \begin{array}{c} dx/dt+6x-5y-t=0 \\ dy/dt+5x-4y-1=0 \\ \end{array} \right. $$
which leads to: $$ \left\{ \begin{array}{c} (D+6)x-5y-t=0 \\ (D-4)y+5x-1=0 \\ \end{array} \right. $$
I multiplied top one by $(D-4)$ and bottom one by $5$ to cancel out ys. In result I got $$ (D+1)^2x=(D-4)t+5 $$ Using repeated root formula, $Xc=c_1e^{-t}+c_2te^{-t}$
Now my problem starts. What should I do with $Dt-4t+5$. All examples in book have something like $e^t$ or $t^2$, which is easy to take derivative of. In my case, would it just be $-4t + 5 + C$?
If so, would trial particular solution be $Xp=At+B$? Do I just ignore "C"?
Thanks!
$$(D+1)^2x=(D-4)t+5$$ $$ (D+1)^2x=1-4t+5$$ So we have $$\implies (D+1)^2x=-4t+6$$
Try this as particular solution $$x_h=At+B$$ $$(D+1)^2(At+B)=-4t+6$$ $$\implies At +2A+B=-4t+6$$ Solve the system to get A and B
Note that the constant C appears when you integrate not when you differentiate...