# Thread: Where am I going wrong in this exact equations problem?

1. ## Where am I going wrong in this exact equations problem?

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

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

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

Then multiplying everything by µ we get $M=e^x(x^2+2x)siny$ and $N=x^2e^xcosy$. Thus, $M_y=e^x(x^2+2x)cosy$ and $N_x=e^x(x^2+2x)cosy$. 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 $x^2e^xsiny+h(x)$

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

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

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

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

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

Then multiplying everything by µ we get $M=e^x(x^2+2x)siny$ and $N=x^2e^xcosy$. Thus, $M_y=e^x(x^2+2x)cosy$ and $N_x=e^x(x^2+2x)cosy$. 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 $x^2e^xsiny+h(x)$

Then I differentiated this with respect to x getting ${\color{red}2xe^x}siny+h'(x)$
Then I compared this to M. $2xe^xsiny+h'(x)=M=e^x(x^2+2x)siny$. So from this I get $h'(x)=x^2e^x$ and then $h(x)=1/3x^3e^x$ making my final answer $x^2e^xsiny+(1/3)x^3e^x=c$.

But according to my book the final answer is only $x^2e^xsiny=c$. I'm not sure where the $(1/3)x^3e^x$ is going. Help please
The part in red above. You need to use the product rule so there's another piece.

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# (x 2)sinydx xcosydy=0 solve using integrating factor

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