Showing that a function is a $3$-parameter family of solutions of a differential equation

Given the differential equation $y''' - y' - e^{2x}\sin^2x = 0$ and $y= c_1+c_2e^x+c_3e^{-x}+(\frac{1}{12}+\frac{9\cos2x - 7\sin2x}{520})e^{2x}$, show that $y$ is a $3$ - parameter family of solutions of the above differential equation.

I've been trying to solve the above problem for about an hour without making much headway. Directly differentiating $y$ was the (only?) strategy I tried, but the resulting derivatives get ugly pretty quickly, despite being able to replace certain terms in terms of previously known terms. And since the differential equation contains a $\sin^2x$ term, I did try replacing the $\cos2x$ and $\sin2x$ terms in terms of $\sin^2x$, but it still didn't help in simplifying the expressions.

Could anyone provide a hint on how to solve this problem? What is the "trick" involved?

• Yes, I double-checked it. This is Problem $5$ from Exercise $4$ (Page $37$) of Ordinary Differential Equations, by Tenenbaum and Pollard. – Train Heartnet Dec 21 '14 at 7:29
• The book hasn't got to explaining how to solve such differential equations yet. So could you explain how you verified it using the derivatives? Did you just directly differentiate $y$ and the subsequent derivatives (with all the ugly terms) or did you use some sort of trick? – Train Heartnet Dec 21 '14 at 7:39
• Oh, I see. Could you provide the derivatives then please, so that I can check my calculations? – Train Heartnet Dec 21 '14 at 7:45

Starting with $$y= c_1+c_2e^x+c_3e^{-x}+(\frac{1}{12}+\frac{9\cos2x - 7\sin2x}{520})e^{2x}$$ and simplifying, you should get $$y'={c_2} e^x-{c_3} e^{-x}+\frac{1}{390} e^{2 x} (-24 \sin (2 x)+3 \cos (2 x)+65)$$ $$y''={c_2} e^x+{c_3} e^{-x}-\frac{1}{195} e^{2 x} (27 \sin (2 x)+21 \cos (2 x)-65)$$ $$y'''={c_2} e^x-{c_3} e^{-x}-\frac{2}{195} e^{2 x} (6 \sin (2 x)+48 \cos (2 x)-65)$$ So $$y'''-y'=\frac{e^{2 x}}{2}-\frac{1}{2} e^{2 x} \cos (2 x)$$