Solve the differential equation yy'sin^2 x + y^2sin xcos x…

Solve the differential equation $yy'\sin^2 x + y^2\sin x\cos x - 1 = 0$.

Explanation of Method

Recognizing $2y\sin^2x\,dy+y^2\sin2x\,dx$ as the exact differential $d(y^2\sin^2x)$ reduces the equation to $d(y^2\sin^2x)=2\,dx$, which integrates directly. The result is verified independently by implicit differentiation with \texttt{sympy}.

Step 1: Differential form

Rearranging: $(1-y^2\sin x\cos x)\,dx-y\sin^2x\,dy=0$.

Step 2: Recognize an exact differential

Since $\sin x\cos x=\tfrac12\sin2x$, multiplying the original equation by $2\,dx$ equivalent gives \[ 2y\sin^2x\,dy+y^2\sin2x\,dx=2\,dx. \] \texttt{sympy} confirms \[ d(y^2\sin^2x)=2y\sin^2x\,dy+2y^2\sin x\cos x\,dx=2y\sin^2x\,dy+y^2\sin2x\,dx, \] so the left side is exactly $d(y^2\sin^2x)$.

Step 3: Integrate

\[ d(y^2\sin^2x)=2\,dx \ \Longrightarrow\ y^2\sin^2x=2x+C. \]

Step 4: Independent verification

Differentiating $y^2\sin^2x=2x+C$ implicitly with \texttt{sp.Function('y')(x)} and \texttt{sympy}, solving for $y'$, and substituting into $yy'\sin^2x+y^2\sin x\cos x-1$ gives a residual that \texttt{sympy.simplify} reduces to $0$ identically, confirming the source's own hand verification.

Result

Final answer:

The source's derivation and self-verification are both confirmed correct by independent \texttt{sympy} check: \[ \boxed{\,y^2\sin^2x=2x+C\,} \] where $C$ is an arbitrary constant (equivalently $y=\pm\sqrt{2x+C}\,\csc x$, for $\sin x\neq0$ and $2x+C\ge0$).