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**steph3824** Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Solve the equation.

$\displaystyle (x+2)sinydx+xcosydy)=0$, µ$\displaystyle (x,y)=xe^x$

So first, $\displaystyle M=(x+2)siny$ and $\displaystyle N=xcosy$. Thus, $\displaystyle M_y=(x+2)cosy$ and $\displaystyle N_x=cosy$

So this is inexact.

Then multiplying everything by µ we get $\displaystyle M=e^x(x^2+2x)siny$ and $\displaystyle N=x^2e^xcosy$. Thus, $\displaystyle M_y=e^x(x^2+2x)cosy$ and $\displaystyle N_x=e^x(x^2+2x)cosy$. So now they are exact

I have chosen to integrate N first since it's a bit easier than M. So integrating N I get $\displaystyle x^2e^xsiny+h(x)$

Then I differentiated this with respect to x getting $\displaystyle {\color{red}2xe^x}siny+h'(x)$

Then I compared this to M. $\displaystyle 2xe^xsiny+h'(x)=M=e^x(x^2+2x)siny$. So from this I get $\displaystyle h'(x)=x^2e^x$ and then $\displaystyle h(x)=1/3x^3e^x$ making my final answer $\displaystyle x^2e^xsiny+(1/3)x^3e^x=c$.

But according to my book the final answer is only $\displaystyle x^2e^xsiny=c$. I'm not sure where the $\displaystyle (1/3)x^3e^x$ is going. Help please