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The differential equation is,

$$\frac{dy}{dx}=x^2-2x^2y$$

I have calculated $μ(x)$ as $μ(x)=e^{2x^3/3}$

thus

$$y=e^{-2x^3/3}\int x^2e^{2x^3/3}\,dx$$

After this I have been having a bit of trouble solving for $y$. I have tried integration by parts but it hasn't helped, probably due to a mistake on my part.

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  • $\begingroup$ All I have to do is evaluate the $y=e^{-2x^3/3}\int x^2e^{2x^3/3}\,dx$ bit though. The standard form was only needed to calculate the integrating factor μ(x) which has been done. $\endgroup$ – copa May 8 '18 at 0:44
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$dy/dx + 2x^2y = x^2$

Then $2x^2$ is P(x) so the integrating factor $v(x) = e^{\int2x^2 dx}$

$= e^{\frac{2}{3}x^3}$

$e^{\frac{2}{3}x^3}(dy/dx + 2x^2y) = e^{\frac{2}{3}x^3}x^2$

$d/dx(e^{\frac{2}{3}x^3}y) = e^{\frac{2}{3}x^3}x^2$......left side is $d/dx(vy)$

$e^{\frac{2}{3}x^3}y = \int e^{\frac{2}{3}x^3}x^2 dx$

$e^{\frac{2}{3}x^3}y = \frac{e^{\frac{2}{3}x^3}}{2} + C$

$y = \frac{1}{2} + \frac{C}{e^{\frac{2}{3}x^3}}$

$y = \frac{1}{2} + Ce^{\frac{-2}{3}x^3}$

As others have pointed out, this equation is separable so this method isn't usual, never mind correct.

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If you are asking how to solve the integral you have, you can use u-substitution.

$u = \frac{2x^3}{3}$

$du = 2x^2$

So

$\int x^2e^{\frac{2x^3}{3}} = \frac{1}{2} \int e^u du = \frac{1}{2} e^{\frac{2x^3}{3}} + C$

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  • $\begingroup$ @randomgirl yes you are correct. Thanks I’ll fix it $\endgroup$ – JBL May 8 '18 at 0:50
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If you're going to integrate by factors, couldn't you just write the question as:

$dy/dx = x^2(1-2y)$

This makes it much easier, because you now can integrate x and y separately, as it's own thing, etc.

$(1-2y)^{-1}\ dy=x^2\ dx$

Integrate both sides to get:

$-\frac12 \ln(1-2y)=x^3/3 + C_1$

$\ln(1-2y) = -\frac{2x^3}3 + C_2$

$1-2y=Ae^{-2x^3/3}$

$y=\frac12(1-Ae^{-2x^3/3})$

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    $\begingroup$ Yes I understand this equation by separation of variables, I just wanted to try it by method of integrating factors. $\endgroup$ – copa May 8 '18 at 0:51

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