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I need to solve the following differential equation:

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

I have tried to solve it using the integrating factor $e^{\int (1/\cos^2x) \mathrm{d}x}$, but things got messed up. How am I to solve it? Should I try a different path?

Thank you!

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    $\begingroup$ Is your problem $\int \frac {dx}{\cos^2(x)}=$ "something" ? $\endgroup$ – Claude Leibovici Aug 12 '14 at 9:37
  • $\begingroup$ I think you have already solved your question. $\endgroup$ – Enthusiastic Engineer Aug 12 '14 at 9:50
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First, solve the homogeneous equation $Y'\cos^2(x)+Y=0$

This leads to $Y=c*e^{\tan(x)}$

Then use the method of variation of the constant : let $y=f(x)*e^{\tan(x)}$

Plug this function into $y'\cos^2(x)+y=\tan(x)$

After simplification, determine $f(x)$ by a simple integration.

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$y'cos^2x+y=tan(x)=y' + y\sec^2x=\tan x \sec^2x$. $$ $$ $I.F. =e^{\int\sec^2xdx}=e^{\tan x}$ $$ $$So the solution becomes : $$ $$ $y* e^{\tan x}=\int\tan x\sec^2 x e^{\tan x} dx$. Now let $tan x=u$. So you get in the RHS $$ $$ $\int ue^udu=e^u(u+1)$. Now resubstitute to get in the RHS $e^{\tan x}(\tan x +1)$. So that the final solution is $$ $$ $y* e^{\tan x}=e^{\tan x}(\tan x +1) + c$ or in a better way $$ $$ $y=ce^{-\tan x} + \tan x +1$

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  • $\begingroup$ Is not the general solution of this D.E, there is problem to dividing by $\cos ^2$ it vanish. $\endgroup$ – Hamou Aug 12 '14 at 10:36

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