Solve the differential equation (x)/(y)=(y')/(x+1)

Solve the differential equation $\dfrac{x}{y}=\dfrac{y'}{x+1}$.

Explanation of Method

Cross-multiply to isolate $y\,dy$ and a polynomial in $x\,dx$, then integrate directly.

\[x(x+1)=y\,y' \quad\Longrightarrow\quad y\,dy=(x^2+x)\,dx.\]

\[\int y\,dy=\int(x^2+x)\,dx \quad\Longrightarrow\quad \frac{y^2}{2}=\frac{x^3}{3}+\frac{x^2}{2}+C.\]

Multiplying by $6$ and relabelling the constant:

\[\boxed{\,3y^2=2x^3+3x^2+c.}\]

Besides the sign condition $2x^3+3x^2+c\ge0$, the original equation is undefined at $y=0$ and $x=-1$, which must be excluded from any solution curve.

Final answer:

\[\boxed{\,3y^2=2x^3+3x^2+c,\qquad c\in\mathbb{R}.}\]