A simple differential equation question

Question

$dy/dx=(xy+2x)/(y-1+x^2)$How to solve this equation?

Answer 1

Hint

You can try integrating factor to make it exact $$ \frac {dy}{dx}=\frac {(xy+2x)}{(y-1+x^2)}$$ $$dy(y-1+x^2)-dx(xy+2x)=0$$ $$-x(y+2)dx +(y-1+x^2)dy=0$$ $$Pdx+Qdy=0$$ The formula for integrating factor depending only on y is $$\frac {d\mu}{\mu dy}=-\frac 1P(\partial_y P-\partial_x Q)$$ Factor integrating is $$ \frac {d\mu}{\mu dy}=-\frac 3 {y+2}$$ $$ \ln|\mu|=-\int \frac 3 {y+2}dy$$

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Another view..

Or consider simply $x$ as a function $$y'=\frac {(xy+2x)}{(y-1+x^2)}$$ $${(y-1+x^2)}=x'x{(y+2)}$$ $${(y-1+x^2)}=\frac 12(x^2)'{(y+2)}$$ Then simply substitute $z=x^2$

$${(y-1+z)}=\frac 12z'{(y+2)}$$ $$z'-\frac {2z}{y+2}=2\frac {(y-1)}{(y+2)}$$ It's easy to solve since its a linear ode