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Show that the differential equation

$(3y^2-x)+2y(y^2-3)y'=0$

admits an integrating factor which is a function of $(x+y^2)$.

Hence solve the equation.

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closed as off-topic by mickep, TZakrevskiy, Chappers, user147263, Winther Sep 28 '15 at 22:41

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    $\begingroup$ Hi and welcome to Math.SE. Please share your own thoughts about the problem. Where do you get stuck? What don't you understand? ... Also, please use mathjax to type your questions, since it will be easier for others to read. Again, welcome! $\endgroup$ – mickep Sep 28 '15 at 19:03
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A sketch:

Assume $I=e^{\int f(x+y^2)d(x+y^2)}$ is an integrating factor of the ODE. (If you understand how the idea of integrating factor work, this is nature to write it in this form, instead of simply $f(x+y^2)$. And I think this is the only tricky thing we need to solve it.)

By definition of integrating factor, $$I \cdot (3y^2-x)dx+I \cdot 2y(y^2-3)dy=0$$ will be exact, i.e. $$\frac{\partial (I \cdot (3y^2-x))}{\partial y}=\frac{\partial (I \cdot 2y(y^2-3))}{\partial x}.$$

Hence we just need to find one possible $f(x+y^2)$. It is straightforward: we only need to evaluate $LHS$ and $RHS$, comparing them, then we will get the requiring $f$, and hence $I$.

Now you can treat the ODE with usual way. If you're still confusing, please let me know.

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