Solve the differential equation (y^2+4sin x)y' - cos x = 0

Solve the differential equation $(y^2+4\sin x)y' - \cos x = 0$.

Step 1: Exactness test and integrating factor

With \(M=\cos x\), \(N=-(y^2+4\sin x)\): \(M_y=0\), \(N_x=-4\cos x\), not exact. But \((N_x-M_y)/M=-4\cos x/\cos x=-4\), a constant, so \(\mu(y)=e^{-4y}\).

Step 2: Multiply through and confirm exactness

\(\tilde M=e^{-4y}\cos x\), \(\tilde N=-e^{-4y}(y^2+4\sin x)\). \texttt{sympy} confirms \(\tilde M_y=\tilde N_x=-4e^{-4y}\cos x\), so the equation is now exact.

Step 3: Integrate \(\tilde M\) with respect to \(x\)

\[ F(x,y)=\int e^{-4y}\cos x\,dx=e^{-4y}\sin x+g(y). \]

Step 4: Determine \(g(y)\), verifying the integration by parts independently

Requiring \(F_y=\tilde N\) gives \(g'(y)=-y^2e^{-4y}\). Calling \texttt{sympy.integrate(y**2*sp.exp(-4*y), y)} independently (not trusting the source's by-hand double integration by parts) returns \[ \int y^2e^{-4y}\,dy=-e^{-4y}\Big(\frac{y^2}{4}+\frac{y}{8}+\frac{1}{32}\Big)+C, \] which matches the source's result exactly. Hence \[ g(y)=-\int y^2e^{-4y}\,dy=e^{-4y}\Big(\frac{y^2}{4}+\frac{y}{8}+\frac{1}{32}\Big), \] and \texttt{sympy} confirms \(g'(y)=-y^2e^{-4y}\) as required.

Step 5: Assemble and simplify the solution

\[ F(x,y)=e^{-4y}\Big(\sin x+\frac{y^2}{4}+\frac{y}{8}+\frac{1}{32}\Big)=C. \] Multiplying by \(e^{4y}\): \[ \sin x+\frac{y^2}{4}+\frac{y}{8}+\frac{1}{32}=Ce^{4y}. \]

Step 6: Independent verification

Implicitly differentiating \(\sin x+y^2/4+y/8+1/32-Ce^{4y}=0\) with \texttt{sympy}, solving for \(y'\), eliminating \(C\) using the same relation, and substituting into \((y^2+4\sin x)y'-\cos x\) gives a residual that \texttt{sympy.simplify} reduces to \(0\) identically.