How to solve a differential equation $(e^x+2\ln y)ydx+xdy=0$

Question

Solve the following differential equation: $$ (e^x+2\ln y)ydx+xdy=0 $$

It is clear that the equation is not exact. So, I tried to express $y'$: $$ \begin{aligned} e^xy+2y\ln y+xy'=0\iff \begin{cases} \left[ \begin{aligned} &y\equiv0\\ &y\equiv e^{-{1/2}} \end{aligned} \right. \ \ \text{if}\ \ x=0\\ y'=-\frac{e^x}{x}\cdot y-\frac{1}{x}\cdot2y\ln y\ \ \ \text{otherwise} \end{cases} \end{aligned} $$ The problem is that the differential equation seems to be non-linear, and I don't know the ways of solving those. Maybe there's an easier way of solving the initial differential equation?

Answer 1

Maybe there's an easier way of solving the initial differential equation? Yes. This equation becomes exact if you mutliply by integrating factor $$\mu (x,y)=\dfrac x y$$

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$$(e^x+2\ln y)ydx+xdy=0$$ Divide by $y$ note that $ d(\ln y )=\dfrac {dy}{y}$ $$(e^x+2\ln y)dx+xd \ln y=0$$ Substitute $u= \ln y$ $$(e^x+2u)dx+xdu=0$$ Multiply by $x$: $$xe^xdx+2uxdx+x^2du=0$$ $$xe^xdx+d(ux^2)=0$$ Integrate. $$xe^x-e^x+x^2 \ln y =C$$