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?

  • $\begingroup$ Yes, I double-checked it. This is Problem $5$ from Exercise $4$ (Page $37$) of Ordinary Differential Equations, by Tenenbaum and Pollard. $\endgroup$ – Train Heartnet Dec 21 '14 at 7:29
  • $\begingroup$ 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? $\endgroup$ – Train Heartnet Dec 21 '14 at 7:39
  • $\begingroup$ Oh, I see. Could you provide the derivatives then please, so that I can check my calculations? $\endgroup$ – Train Heartnet Dec 21 '14 at 7:45

As Amzoti commented, compute each derivative and simplify it as much as you can before going to the next derivative (otherwise, it could be a nightmare).

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 $$y'''-y'=\frac{e^{2 x}}{2}-\frac{1}{2} e^{2 x} \cos (2 x)$$

I am sure that you can take from here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.