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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

2 Answers 2

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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 $$u(x,y)=\int m.dx=\int (ye^{xy} + cos(x))dx=\\e^{xy}+sinx +g(y) \to $$ $$u_y=x.e^{xy}+0+g'(y) =xe^{xy} \\\to g'(y)=0 \to g(y)=c\\so \\ u(x,y)=e^{xy}+sinx+c$$

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    Oh gee, thank you. For some inane reason, I was skipping past the product rule. Thank you – 2017-01-30
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    Can you go on ? – 2017-01-30
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    So, I saw that mistake, but I am having trouble finding the integrals based off of m. and n. I am not sure how to find the integral of (xe^(xy)) or (ye^(xy)) that will lead to some general solution psi = c. And sorry my formatting is so terrible. I am not competent enough to format my equations like you. @Khosrotash – 2017-01-30
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    @BryanChen:I complete it .see and find your mistake – 2017-01-30
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    Thank you, I made a silly error. I was accidentally integrating m with respect to y and n with respect to x. Maybe doing math in the wee hours of the night is to blame. Thanks for your help again. I posted another question about difference equations, and I am wondering if you could reformat that one like you did for this. Do not worry if you can't, you have helped me enough already. Thanks @Khosrotash – 2017-01-30
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\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.