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The equation is as follows: $$(\cos(x)e^x+xy+4y^2-y)dx+(-x-8y)dy=0$$

I have worked out that it isn't exact and I have worked out the integrating factor to be $e^x$

I have then multiplied the original equation by $e^x$ to get $$(\cos (x)e^{2x}+xye^x+4y^2e^x-ye^x)dx+(-xe^x-8ye^x)dy=0$$

I am stuck trying to find the solution in terms of $f(x,y)=c$

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  • $\begingroup$ Do you mean $\cos(xe^x)$ or $\cos(x)e^x$? $\endgroup$ Mar 17 '20 at 5:19
  • $\begingroup$ @callculus its $e^x \cos x$ $\endgroup$ Mar 17 '20 at 5:55
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    $\begingroup$ Look at the line where OP multiplied the DE with integrating factor @callculus $\endgroup$ Mar 17 '20 at 6:00
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    $\begingroup$ @LostInSpace Good catch. $\endgroup$ Mar 17 '20 at 6:01
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    $\begingroup$ lol @callculus Thank you ..... $\endgroup$ Mar 17 '20 at 6:01
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$$(\cos xe^x+xy+4y^2-y)dx+(-x-8y)dy=0$$ Rearrange some terms: $$\cos xe^xdx-(d(xy)-xydx)+(4y^2dx-8ydy)=0$$ Factor integrating is $e^{-x}$ not $e^x$: $$\cos xdx-d(e^{-x}xy)-e^{-x}(-4y^2dx+8ydy)=0$$ $$\cos xdx-d(e^{-x}xy)-4d(e^{-x}y^2)=0$$ Integrate. $$\sin x-e^{-x}xy-4e^{-x}y^2=C$$

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Let it be $Mdx+Ndy=0, M=\cos x e^x+xy+4y^2-y, N=-x-8y$ so Here $$\frac {\partial M}{\partial y}=x+8y-1, \frac{\partial N}{\partial x}=-1=$$ Integrating factor $$\mu(x)= \exp ~[\int \frac{[\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}] dx}{N}] =e^{-x}$$ So the soluytion of the ODE can be written as $$\int e^{-x} (\cos x e^x+xy+4y^2-y) dx \text{[treat $y$ as constant]}+ \int 0 dy=C$$ $$\implies \sin x-e^{-x}y(x+4y)=C$$

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  • $\begingroup$ Thankyou, I didn't realise I had miscalculated my integrating factor. This has helped me a lot $\endgroup$
    – Kat L
    Mar 17 '20 at 5:52

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