Hints on how to solve $(x+y^2)dy = ydx$?

Question

I'm looking for hints on how to solve the differential equation: $(x+y^2)dy = ydx$ . I tried finding an integrating factor and dividing both sides by $y$ but that didn't work.

Answer 1

In $(x+y^2)dy=ydx$ you can indeed divide by $y^2$ to get $$ d\left(\frac{x}{y}\right)=\frac{y\,dx-x\,dy}{y^2}=dy $$ which is directly integrable.

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For the first version of the question, $(x^2+y^2)dy=ydx$, observe that $$ \frac{dx}{dy}=y+\frac{x^2}{y} $$ looks like a Riccati equation. Set $x(y)=-\dfrac{yu'(y)}{u(y)}$ to get a second order linear ODE in $u$ $$ y+\frac{yu'^2}{u^2}=x'=-\frac{yu''+u'}{u}+\frac{yu'^2}{u^2} \\\iff\\ yu''+u'+yu=0. $$ Then apply power series expansion or identify the special function type that solves this equation.

Answer 2

Hint

Write the equation as $$x+y^2=y x'\implies yx'-x=y^2$$ which is quite simple to solve.

Answer 3

Consider x' instead of y'it's simplier

$$(x+y^2)dy = ydx$$ $$(x+y^2) = y\frac {dx}{dy}$$ $$\frac {x'y-x}{y^2}=1$$ $$(\frac xy)'=1$$ Simply integrate now $$\frac xy=\int dy=y+K$$ $$\boxed{x(y)=y^2+Ky}$$