Please help solving this inital-value problem

Question

The initial-value problem

$$\begin{cases}(ye^{xy} + cos(x))dx + (xe^{xy})dy = 0\\y(\dfrac{\pi}{2}) = 0\end{cases}$$

From my calculations it seems the integrating factor depends on y and x, but I am unsure how to find the correct integrating factor.

Please help

Answer 1

Hint:$$m.dx+n.dy=0 \\$$ $$\begin{cases}(ye^{xy} + cos(x))dx + (xe^{xy})dy = 0\\y(\dfrac{\pi}{2}) = 0\end{cases}\\ \frac{\partial (ye^{xy} + cos(x)) }{\partial y}=1.e^{xy}+xy.e^{xy}+0\\ \frac{\partial (xe^{xy})}{\partial x}=1.e^{xy}+yxe^{xy}\\so \\ \frac{\partial n}{\partial x}=\frac{\partial m }{\partial y}\\exact \space equation$$now $$u(x,y)=\int m.dx=\int (ye^{xy} + cos(x))dx=\\e^{xy}+sinx +g(y) \to $$ $$u_y=x.e^{xy}+0+g'(y) =xe^{xy} \\\to g'(y)=0 \to g(y)=c\\so \\ u(x,y)=e^{xy}+sinx+c$$

Answer 2

\begin{eqnarray} \big(ye^{xy} + \cos x\big)dx + (xe^{xy})dy &=& 0\\ e^{xy}(ydx+xdy) + \cos x dx &=& 0\\ e^{xy}d(xy) + \cos x dx &=& 0\\ d(e^{xy}) + d(\sin x)&=& 0\\ d(e^{xy}\sin x)&=& 0\\ e^{xy}\sin x&=& C\\ \end{eqnarray} with $y(\dfrac{\pi}{2}) = 0$ we have $e^{\frac{\pi}{2}0}\sin\dfrac{\pi}{2}=C$ or$C=1$ and $\color{blue}{e^{xy}\sin x=1}$ is final solution.