I have the following DE:
$$(x+ye^{y/x})dx - xe^{y/x} dy = 0$$
I want to sub in $ u = \frac yx $ . When I solve for and take the derivative of $y=ux$, I get $dy = xdu + u$. Is this right? I am also wondering if I can write it as $\frac {dy}{dx} = x\frac {du}{dx} + u$? My last question is could I have solved for x and solved for $dx$?
Thank you.
$$(x+ye^{y/x})dx - xe^{y/x} dy = 0$$ $$(x+ye^{y/x}) = xe^{y/x}y'$$ $$(\frac y x+e^{-y/x}) = y'$$ $$y'=(\frac y x+e^{-y/x})$$ substitute $y=tx$ and $y'=t'x+t$ $$t'x+t=t+e^{-t}$$ $$t'x=e^{-t}$$ It's seperable now $$\int e^t dt=\int \frac {dx} x=\ln(x)+K$$ $$ e^t =\ln(x)+K$$ $$ e^{y/x} =\ln(x)+K \implies e^y=(\ln(x)+K)^x$$ $$y=\ln(\ln(x)+K)^x=x\ln(\ln(x)+K)$$ $$\boxed{y(x)=x\ln(\ln(x)+K)}$$