Solve the differential equation [sin(x+y) + xcos(x+y)] dx +…

Solve the differential equation $[\sin(x+y) + x\cos(x+y)]\,dx + x\cos(x+y)\,dy = 0$.

Substitute \(u = x+y\)

Let \(u = x+y\), so \(dy = du - dx\). The equation becomes

\[[\sin u + x\cos u]\,dx + x\cos u\,(du - dx) = 0.\]

Simplify

Expanding, the \(x\cos u\,dx\) terms cancel, leaving

\[\sin u\,dx + x\cos u\,du = 0.\]

Separate variables

\[\frac{dx}{x} = -\frac{\cos u}{\sin u}\,du = -\cot u\,du.\]

Integrate both sides

\[\int\frac{dx}{x} = -\int\cot u\,du \quad\Longrightarrow\quad \ln|x| = -\ln|\sin u| + C_1,\]

so \(\ln|x\sin u| = C_1\), giving \(x\sin u = C\).

Back-substitute \(u = x+y\)

\[x\sin(x+y) = C.\]

Explicitly, \(y = \arcsin(C/x) - x\) for \(|C/x|\le 1\).

Verify by implicit differentiation and exactness

Differentiating \(x\sin(x+y) = C\) implicitly reproduces the original equation exactly:

\[[\sin(x+y) + x\cos(x+y)]\,dx + x\cos(x+y)\,dy = 0.\]

Independently, with \(M = \sin(x+y)+x\cos(x+y)\) and \(N = x\cos(x+y)\), direct computation gives \(\partial M/\partial y = \partial N/\partial x = \cos(x+y) - x\sin(x+y)\), so the original equation is itself exact. A sympy check confirms both the implicit-differentiation identity and the exactness condition hold identically.