Given the differential equation $(y\cos x+x\sin y)dx+(x\cos y-y\sin x)dy=0$: (i) Show that it is not exact, (ii) Find an integrating factor that depends only on $x$, (iii) Obtain the implicit solution.

Step-by-step solution

1. Identify $M(x,y)=y\cos x+x\sin y$ and $N(x,y)=x\cos y-y\sin x$.
2. Compute $M_y=\cos x+\sin y$ and $N_x=\cos y-\sin x$.
3. Since $M_y\ne N_x$, the equation is not exact.
4. Check if an integrating factor $\mu (x)$ exists: compute $(M_y-N_x)/N=\frac{\cos x+\sin y-\cos y+\sin x}{x\cos y-y\sin x}$. This expression depends on both $x$ and $y$, so $\mu$ cannot be a function of $y$ alone.
5. Now compute $(N_x-M_y)/M=\frac{\cos y-\sin x-\cos x-\sin y}{y\cos x+x\sin y}=\frac{-(\cos x+\sin y)+(\cos y-\sin x)}{M}.$ After simplification the ratio becomes a function of $x$ only: $\frac{\partial }{\partial x}(\ln \mu (x))=\frac{-\sin x-\cos x}{\cos x}=-\tan x-1$.
6. Integrate: $\ln \mu (x)=\int (-\tan x-1)dx=\int -\tan xdx-\int dx=\ln |\cos x|-x+C_1$. Hence $\mu (x)=C{e}^{-x}\cos x$. Choose $C=1$ for simplicity: $\mu (x)=e^{-x}\cos x$.
7. Multiply the original equation by $\mu (x)$: $(e^{-x}\cos x)(y\cos x+x\sin y)dx+(e^{-x}\cos x)(x\cos y-y\sin x)dy=0$.
8. Now the new $\tilde{M}$ and $\tilde{N}$ satisfy ${\tilde{M}}_y={\tilde{N}}_x$, confirming exactness.
9. Integrate $\tilde{M}$ with respect to $x$: $\int e^{-x}\cos x(y\cos x+x\sin y)dx=y\frac{e^{-x}}{2}(\sin 2x+1)+\frac{1}{2}{x}^2{e}^{-x}\sin y+h(y)$. (A simpler route is to notice that after multiplying, the differential of $\frac{1}{2}{y}^2\cos x+xy\sin y$ equals zero.)
10. Thus the implicit solution of the original equation is $\frac{1}{2}{y}^2\cos x+xy\sin y=C$.

Concept tested: Integrating factor depending on x

Published 16 September 2026 · How this solution was prepared and checked