Find the general solution of the differential equation whose…

Find the general solution of the differential equation whose solution is given by $y^2+cx=x^3$.

Explanation of Method

Differentiate to eliminate $c$, obtain a first-order ODE, then integrate it back using the same elimination relation to recover the original family.

Step 1: Eliminate $c$

Differentiating $y^2+cx=x^3$ gives $2yy'+c=3x^2$. Since $c=x^3-y^2$ from the original relation,

\[\boxed{\,2yy'=y^2+3x^2-x^3,\qquad y\neq0.}\]

Step 2: Recover the family

From $2yy'+c=3x^2$, integrate directly: $\displaystyle\int 2y\,\frac{dy}{dx}\,dx=\int(3x^2-c)\,dx$, giving $y^2=x^3-cx+C_0$. Taking $C_0=0$ (absorbed into $c$) reproduces

\[y^2+cx=x^3.\]

Result

Final answer:

\[\boxed{\,2yy'=y^2+3x^2-x^3\ \Longleftrightarrow\ y^2+cx=x^3,\qquad y=\pm\sqrt{x^3-cx}.}\]