What is the particular integral?

Question

If $y=\varphi(x) $ is a particular solution of $y''+(\sin(x))y'+2y=e^x$ and $\psi(x)$ is a particular solution of $y''+(\sin(x))y'+2y=\cos(2x)$,then what is the particular solution of $y''+(\sin(x))y'+2y=e^x+2\sin^2(x)$?

Since $LHS$ of equation is same while their $RHS$ are different.Since Particular integral is obtained with aid of the function on $RHS$,so i tried to find the relation among the $RHS $ terms of each equation$(e^x+2\sin^2(x)=e^x+1-\cos2x)$,$RHS $ of $3^{rd}$=$RHS $ of $1^{st}-RHS $ of $2^{nd}$

What to do next?

Answer 1

It is given that:

$\phi'' + \sin x \phi + 2\phi = e^x\\ \psi'' + \sin x \psi + 2\psi = \cos 2x$

You have that:

$\cos 2x = 1-2\sin^2 x$

Differentiation is linear:

i.e. $(u+v)' = u' + v'$

$(\phi - \psi)'' + \sin x (\phi-\psi)' + 2(\phi - \psi) = e^x - \cos 2x = e^x + 2\sin x - 1\\ (\phi - \psi+\frac 12)'' + \sin x (\phi-\psi+\frac 12)' + 2(\phi - \psi+\frac 12) = e^x + 2\sin x$

Answer 2

Hint

Your problem is the contant 1 for that just try $y_p=K$ into the equation to get the value of the particular solution of $$y_p''+(\sin(x))y_p'+2y_p=1$$ $$\implies k''+\sin(x)K'+2K=1 \implies K=1/2$$ Then substract $\phi(x)$ and $\psi(x)$ to get $e^x-\cos(2x)$