Solve the differential equation y^2y' + x^2sin^3 x = y^3cot x

Solve the differential equation $y^2y' + x^2\sin^3 x = y^3\cot x$.

Explanation of Method

Dividing by $y^2$ exposes a Bernoulli equation with $n=-2$. The substitution $v=y^3$ linearizes it, and the resulting linear equation is solved with integrating factor $1/\sin^3x$.

Rewrite as a Bernoulli equation

Dividing by $y^2$: \[\frac{dy}{dx}-y\cot x=-\frac{x^2\sin^3x}{y^2}.\] This is Bernoulli with $n=-2$.

Substitute $v=y^3$

With $v=y^{1-(-2)}=y^3$, $dv/dx=3y^2\,dy/dx$. Multiplying the equation by $3y^2$ and using $v=y^3$: \[\frac{dv}{dx}-3v\cot x=-3x^2\sin^3x.\]

Solve the linear equation in $v$

The integrating factor is \[\mu(x)=e^{\int-3\cot x\,dx}=e^{-3\ln|\sin x|}=\frac{1}{\sin^3x}.\] Multiplying through, the left side becomes $\dfrac{d}{dx}\!\left(\dfrac{v}{\sin^3x}\right)=-3x^2$.

Integrate

\[\frac{v}{\sin^3x}=-x^3+C\ \Longrightarrow\ v=(C-x^3)\sin^3x.\]

Back-substitute and verify

Since $v=y^3$, $y^3=(C-x^3)\sin^3x$. Differentiating this and substituting back reproduces $y^2y'+x^2\sin^3x-y^3\cot x\equiv0$ exactly (confirmed symbolically).

Result

Final answer:

Verified as correct. The general solution is:

\[\boxed{\,y^3=(C-x^3)\sin^3x\,}\]

equivalently $y=(C-x^3)^{1/3}\sin x$, with $C$ an arbitrary constant.