The equation (xcos y-ysin y) dx+(xsin y+ycos y) dy=0 is…
The equation $(x\cos y-y\sin y)\,dx+(x\sin y+y\cos y)\,dy=0$ is claimed to have an integrating factor of the form $\mu=\mu(x)$ or $\mu=\mu(y)$. Test this claim rigorously and find the general integral if such a factor exists.
Explanation of Method
Recompute $M_y$ and $N_x$ directly (they are unequal, confirming the equation is genuinely not exact). Then prove — analytically, not just by trial guesses — that the defining ratios for $\mu(x)$ and $\mu(y)$ each depend on both $x$ and $y$, so neither factor can exist; a broader symbolic sweep (powers of $x,y$, functions of $xy,\,x\pm y,\,x^2+y^2,\,x/y$, and $1/M,\,1/N$) confirms no simple closed-form integrating factor exists at all. The claim in the problem statement is therefore false as posed, most likely due to a transcription error.
Step 1: Recompute $M_y$ and $N_x$ directly
With $M=x\cos y-y\sin y$ and $N=x\sin y+y\cos y$:
\[M_y=-x\sin y-(\sin y+y\cos y)=-x\sin y-\sin y-y\cos y,\qquad N_x=\sin y.\]
These are confirmed unequal by direct symbolic differentiation (sympy), so the equation is genuinely not exact — the source's exactness computation itself was correct.
Step 2: Prove no $\mu(x)$ can exist
\[\frac{M_y-N_x}{N}=\frac{-x\sin y-2\sin y-y\cos y}{x\sin y+y\cos y}=-1-\frac{2\sin y}{N}.\]
Symbolic differentiation of this ratio with respect to $y$ gives a nonzero expression identically (not merely at isolated points), so this ratio provably depends on $y$ as well as $x$. Hence no integrating factor $\mu=\mu(x)$ can exist — this is a proof, not a failed guess.
Step 3: Prove no $\mu(y)$ can exist
\[\frac{N_x-M_y}{M}=\frac{x\sin y+2\sin y+y\cos y}{x\cos y-y\sin y}.\]
Symbolic differentiation with respect to $x$ gives $-\dfrac{y+\sin 2y}{M^2}\not\equiv0$, so this ratio provably depends on $x$ as well as $y$. Hence no integrating factor $\mu=\mu(y)$ can exist either.
Step 4: Broader search rules out other simple factors
A systematic sympy sweep over $\mu\in\{x,\,y,\,xy,\,1/x,\,1/x^2,\,1/y,\,1/y^2,\,1/(x^2+y^2),\,1/M,\,1/N\}$, and over $\mu=\mu(s)$ for $s=x+y,\,x-y,\,xy,\,x^2+y^2,\,x/y$, shows every candidate fails the exactness test after multiplication (all of these were also tried, inconclusively or with hand errors, in the source). Direct attempted integration of $M$ in $x$ (holding $y$ fixed) and matching against $N$ likewise leaves a residual term that depends on $y$, confirming no potential function exists for the equation as stated.
Step 5: Diagnose the source's final step
Rather than reporting this negative result, the source's write-up abandons its own (correct) diagnostic work and fabricates a final expression $F=xy\cos y$ with the placeholder "$+C_{\text{terms}}$" that was never derived. Checking it directly: $F_x=y\cos y\ne M$ and $F_y=x\cos y-xy\sin y\ne N$, so this boxed "answer" does not satisfy the equation at all — it is not a valid solution, merely an unverified guess presented as one.
Result
Final answer:
\[\boxed{\text{No integrating factor }\mu=\mu(x)\text{ or }\mu=\mu(y)\text{ exists for this equation as stated.}}\]
This is proved rigorously (not by exhausting guesses): $(M_y-N_x)/N$ and $(N_x-M_y)/M$ are each shown by symbolic differentiation to depend on both $x$ and $y$, which is incompatible with either single-variable integrating-factor condition. A further sweep over common two-variable forms ($xy$, $x\pm y$, $x^2+y^2$, $x/y$, $1/M$, $1/N$) also fails. The problem statement's premise is therefore not satisfiable for the equation as transcribed, indicating a typo in the original source (as the source itself suspected) rather than a solvable exact/integrating-factor problem. The source's own diagnostic steps ruling out $\mu(x)$, $\mu(y)$, and the guesses $1/y,\,x,\,y,\,1/y^2,\,xy$ were essentially correct; its error was fabricating an unverified final answer ($F=xy\cos y$) instead of reporting this conclusion — that fabricated answer fails both $F_x=M$ and $F_y=N$ when checked directly.