A second order linear, non - homogeneous ODE has a form of [tex]ay''+by'+cy=g(x)[/tex]
The general solution to, [tex]a(x)y''+b(x)y'+c(x)y=0[/tex]
Can be written as,
[tex]y=y_h+y_p[/tex]
Where [tex]y_h[/tex] is a solution to the homogeneous ODE and [tex]y_p[/tex] the particular solution, function that satisfies the non - homogeneous equation.
We can solve [tex]y_h[/tex] by rewriting the equation,
[tex]ay''+by'+cy=0\Longrightarrow(e^{xy})''+4e^{xy}=0[/tex]
Which simplifies to,
[tex]e^{xy}(y^2+4)=0[/tex]
From here we get two solutions,
[tex]y_{h1}=2i, y_{h2}=-2i[/tex]
So the form here refines,
[tex]y_h=c_1\cos(2x)+c_2\sin(2x)[/tex]
The same thing we do with [tex]y_p[/tex] to get form of,
[tex]y_p=-2x\cos(2x)[/tex]
From here the final form emerges,
[tex]y=\boxed{c_1\cos(2x)+c_2\sin(2x)-2x\cos(2x)}[/tex]
Hope this helps.
r3t40
