Find a general solution to the ODE $y'' - 3y' +2y = 10\sin(x) + 2\cos(2x)$

Question

Hello I've been trying to do this question for a bit but I'm not too sure how to proceed or apply the concept to this question. Any clues or hints please? Thank you!

Answer 1

the general solution is $$y=y_h+y_c$$ The homogeneous solution: $$r^2-3r+2=0$$ $$r=1,r=2$$ so $$y_h=c_1e^x+c_2e^{2x}$$ Complementary Solution :

let $$y_c=A\sin x+B\cos x+C\sin2x+D\cos2x$$ now drive it two times and substitute the equations in the original differential equation to get the constants

Answer 2

HINT

I would start by noticing that \begin{align*} y'' - 3y' + 2y = 10\sin(x) + 2\cos(2x) & \Longleftrightarrow (y'' - y') - 2(y' - y) = 10\sin(x) + 2\cos(2x)\\\\ & \Longleftrightarrow (y' - y)' - 2(y' - y) = 10\sin(x) + 2\cos(2x)\\\\ & \Longleftrightarrow z' - 2z = 10\sin(x) + 2\cos(2x)\\\\ & \Longleftrightarrow (e^{-2x}z)' = 10e^{-2x}\sin(x) + 2e^{-2x}\cos(2x) \end{align*} Then you integrate both sides and apply the integrating factor once again.

Can you take it from here?