Find the differential equation whose general solution is the…

Find the differential equation whose general solution is the one-parameter family $x^2+c\,y^2=2y$, and show that the family is recovered by solving that ODE.

Step 1: Differentiate and eliminate $c$

Implicit differentiation gives $2x+2cyy'=2y'$, i.e. $x+y'(cy-1)=0$. Since $c=\dfrac{2y-x^2}{y^2}$, we get $cy-1=\dfrac{y-x^2}{y}$, and multiplying by $y$ yields the first-order ODE

\[\boxed{\,(y-x^2)y'+xy=0.}\]

Step 2: Solve with $y=x^2u$

Setting $u=y/x^2$ transforms the equation (for $x\neq0$) into $x(u-1)u'+u(2u-1)=0$, which separates as

\[\frac{u-1}{u(2u-1)}\,du=-\frac{dx}{x}.\]

Partial fractions give $\dfrac{u-1}{u(2u-1)}=\dfrac1u-\dfrac{1}{2u-1}$, so integrating and exponentiating produces $\dfrac{u^2}{2u-1}=\dfrac{K^2}{x^2}$.

Step 3: Back-substitute

Replacing $u=y/x^2$ and simplifying gives $y^2=K^2(2y-x^2)$, i.e. $x^2+\dfrac{1}{K^2}y^2=2y$. Renaming $c=1/K^2$ recovers exactly $x^2+cy^2=2y$. The case $x=0$ is checked separately from the original algebraic relation.