Solution (a) Let $M=\cos y-\cos x$ and $N=e^{y}-x\sin y$. Then $$\begin{equation*} \frac{\partial M}{\partial y}=-\sin y\qquad \hbox{and}\qquad \frac{\partial N}{\partial x}=-\sin y \end{equation*}$$

Since $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$ for all $x$ and $y$, the differential equation is exact over the $xy$-plane.

(b) Since the differential equation is exact, there is a function $z=$ $f(x,y)$ so that $$\begin{equation*} \frac{\partial f}{\partial x}=M=\cos y-\cos x\qquad \hbox{and}\qquad \frac{\partial f}{\partial y}=N=e^{y}-x\sin y \end{equation*}$$

We integrate $\dfrac{\partial f}{\partial x}=M=\cos y-\cos x$ partially with respect to $x$ (holding $y$ constant). Then $$\begin{equation*} f(x,y)=\int ( \cos y-\cos x)\,dx=x\cos y-\sin x+B(y) \tag{3} \end{equation*}$$

where $B,$ which is yet unknown, is a function of $y$ alone. To find $B,$ we differentiate $f$ with respect to $y$. $$\begin{equation*} \frac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}\left[ x\cos y-\sin x+B(y)\right] =-x\sin y-0+\dfrac{\partial }{\partial y}B(y)=-x\sin y+ \dfrac{\it dB}{dy} \end{equation*}$$

Since $\dfrac{\partial f}{\partial y}=N=e^{y}-x\sin y,$ we have $$\begin{eqnarray*} -x\sin y+\dfrac{\it dB}{dy} &=&e^{y}-x\sin y \\ \dfrac{\it dB}{dy} &=&e^{y}\\ B(y)&=&e^{y}+C \end{eqnarray*}$$

where $C$ is a constant. Now we substitute $B(y)$ into (3) to obtain the general solution $$f(x,y)=x\cos y-\sin x+e^{y}+C=0$$

To find the particular solution that satisfies the boundary condition $y(\pi )=0$, we need to find $C$ so that $y=0$ when $x=\pi .$ Then $$\begin{eqnarray*} \pi \cos 0-\sin \pi +e^{0}+C& =&0 \\[3pt] \pi +1+C& =&0 \\[3pt] C& =&-( 1+\pi) \end{eqnarray*}$$

The particular solution is $\ x\cos y-\sin x+e^{y}=\pi +1.$