Solve differential equation using integrating factor $$y'-y/x= \tan(y/x)$$
I get the integrating factor to be $1/x$
Multiply that out and the LHS $= d/dx(y (1/x))$ and the RHS then becomes $\tan(y/x) (1/x)$
Don't know where to go from that.
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Sign up to join this communitySolve differential equation using integrating factor $$y'-y/x= \tan(y/x)$$
I get the integrating factor to be $1/x$
Multiply that out and the LHS $= d/dx(y (1/x))$ and the RHS then becomes $\tan(y/x) (1/x)$
Don't know where to go from that.
Don't agree with 1/x as integrating factor. Following Chappers idea works perfect.
$\frac{\text{dy}}{\text{dx}}=\frac{y}{x}+\tan \left(\frac{y}{x}\right)$
Substitude:
$\frac{y}{x}=u$
$y=x u$
$\frac{\text{dy}}{\text{dx}}=x\frac{\text{du} }{\text{dx}}+u$
$ x\frac{\text{du}}{\text{dx}}+u=u+\tan (u)$
$ x\frac{\text{du}}{\text{dx}}=\tan (u)$
$\frac{\text{du}}{\tan (u)}=\frac{\text{dx}}{x}$
Now on both sides logarithmic derivatives:
$cos (u)\frac{\text{du} }{\sin (u)}=\frac{\text{dx}}{x}$
$\frac{d}{\text{du}}(\ln \sin (u))=\frac{d}{\text{dx}}(\ln (x))$
$\ln (\sin (u))=c+\ln (x)$
Solving y should be easy: $\sin (u)=C x$
$u=\arcsin (C x)$
$y=x \arcsin (C x)$
In my opinion, conditions for integrating factor are hard to solve.