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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!

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    Any means of solving this is going to be equivalent to integrating the $z$ that you found. So yes, you're going about it the right way; but maybe you can simplify the expression that you found for $z$.2012-07-01

1 Answers 1

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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

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

$-2p\tan x=p'$

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

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

$p=\cos^2x$

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

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