I have the following ODE:

$$y'' - 2y'\tan(x)-y=\sin(x)$$

I am at a loss where to start. All the methods described in my textbook assume knowledge of the complementary function to solve $2$nd order ODEs with variable coefficients. However in a case like this, it seems that the roots of the related homogeneous ODE, $$y'' - 2y'\tan(x)-y=0 ,$$ cannot be found.

Could anyone give me a hint on how to start tackling this problem? Or what methods to use?


I don't know a general way of solving differential equations like this, but if you desperately multiply your equation with $\cos (x)$ to get a trigonometric function as a factor in all terms, $$ \cos (x)y''-2\sin (x)y'-\cos (x) y=\sin x\cos x, $$ then you might recognize the left-hand side as $$ (\cos (x) y)'', $$ so your differential equation is really $$ (\cos (x)y)''=\sin x\cos x. $$ Now integrate twice.


Hint Some experimentation with substitutions of the form $y(x) = u(x) v(x)$ (for a fixed function $v$) aimed at eliminating the first-order term shows that substituting $$y(x) := \frac{u(x)}{\cos x}$$ simplifies the differential expression on the l.h.s. of the equation to $$\frac{u''(x)}{\cos x},$$ and so rearranging gives the entirely tractable equation $$u''(x) = \sin x \cos x$$ in $u(x)$: Since only the second derivative of $u$ appears, this amounts to two (easy) integrations.

  • $\begingroup$ Oh, I realize I added a very similar solution to yours. Since our first viewpoint is a bit different, I will let my answer stand, but you were first (↑) $\endgroup$ – mickep Oct 11 '15 at 17:31
  • $\begingroup$ @mickep Yes, do leave it, it motivates the change of variable very nicely, and probably better than my answer does. (+1) $\endgroup$ – Travis Willse Oct 11 '15 at 17:58

show that $$\frac{1}{\cos(x)}$$ and $$\frac{x}{\cos(x)}$$ are solutions

  • 1
    $\begingroup$ Now is that a method to solve differential equations? Just to verify given solutions? $\endgroup$ – mickep Oct 11 '15 at 17:30
  • $\begingroup$ i don't think that there is a general method $\endgroup$ – Dr. Sonnhard Graubner Oct 11 '15 at 17:31

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.