I'd really love your help with solving we following differential equation $$y''-2y'\tan x=\frac{1}{\cos^3x}.$$

First I tried to do it with $z=y'$ but it's just impossible,$z$ is a big and not nice expression, and to integrate it would be very hard problem. Then I though of Euler equations, but it's not in the correct form for doing it.

What should I do?

Thanks a lot!

  • $\begingroup$ Can you solve the homogenous equation (the same LHS equals $0$)? $\endgroup$ – Did Jul 1 '12 at 7:47
  • $\begingroup$ Correct. That's what I get, but then $z= \ln (\frac{\sin(t/2)+\cos(t/2)}{\cos(t/2) -\sin(t/2)}) \cdot \cos^{-2}t +c_1 \cdot \cos^{-2}t$ Integrate this is a difficult job. $\endgroup$ – Jozef Jul 1 '12 at 7:52
  • $\begingroup$ Every detail on this step should be included in your post. $\endgroup$ – Did Jul 1 '12 at 8:04
  • $\begingroup$ @Jozef Yes, it is a difficult job, but it can be done. By the way, WolframAlpha solves the equation exactly like that. $\endgroup$ – Siminore Jul 1 '12 at 8:18
  • $\begingroup$ @Siminore: so? and yes, I find it very difficult in a test to integrate $z$. $\endgroup$ – Jozef Jul 1 '12 at 8:24

I'm not sure it's as hard as you're making it out to be. First, let's find an integrating factor. We want

$$py''-2py'\tan x=(py')'$$

$$-2p\tan x=p'$$

$$\frac p{p'}=-2\frac{\sin x}{\cos x}$$

$$\ln p=2\ln(\cos x)=\ln(\cos^2x)$$


Multiplying through by our integrating factor, we get

$$y''\cos^2x-2y'\sin x\cos x=(y'\cos^2x)'=\sec x$$

$$y'\cos^2x=\ln(\sec x+\tan x)+C$$

$$y'=\sec^2x\ln(\sec x+\tan x)+C\sec^2x$$

$$y=\int\sec^2x\ln(\sec x+\tan x)dx+C\int\sec^2xdx$$

The second part is simply $C\tan x$. For the first integral, we'll use integration by parts. Obviously, we want that logarithm to go away, so that's the part we'll take the derivative of.

$$u=\ln(\sec x+\tan x),du=\sec x$$

$$dv=\sec^2xdx,v=\tan x$$

$$\int\sec^2x\ln(\sec x+\tan x)dx=\tan x\ln(\sec x +\tan x)-\int\sec x\tan xdx=$$

$$\tan x\ln(\sec x+\tan x)-\sec x$$

Putting everything together, we have

$$y=\tan x\ln(\sec x+\tan x)-\sec x+k_1\tan x+k_2$$


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.