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The ODE is

($xy^{3} + x^{2}y^{7}) \frac{dy}{dx} = 1$

I have tried everything like integrating factor,it is not homogenous and not linear differential equation..What should be done now?

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Re-arranging your differential equation, we have $$\frac{dx}{dy}=xy^3+x^2y^7$$ $$\frac{dx}{dy}-xy^3=x^2y^7$$ $$\frac{1}{x^2}\cdot \frac{dx}{dy}-\frac{1}{x}\cdot y^3=y^7$$ $$-\frac{d}{dy}\left(\frac{1}{x}\right)-y^3\cdot \left(\frac{1}{x}\right) =y^7$$

Put $u=\frac{1}{x}$

You get $$\frac{du}{dy}+uy^3=y^7$$

The integrating factor comes out to be $e^\frac{y^4}{4}$.

Then we have $$\frac{d}{dy}\left(ue^\frac{y^4}{4}\right)=y^7e^\frac{y^4}{4}$$

Now $$\int y^7e^\frac{y^4}{4} dy = \int y^4 e^\frac{y^4}{4} \cdot y^3 dy$$ $$=\int 4ze^z dz$$ where $z=\frac{y^4}{4}$.

Can you complete the integration now?

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  • $\begingroup$ I try to integrate it,But integrating factor is insane.Can you provide me details on how to solve it $\endgroup$ – Taylor Ted Dec 31 '15 at 5:28
  • $\begingroup$ @TaylorTed I have added a little more hint. See if it helps now. $\endgroup$ – SchrodingersCat Dec 31 '15 at 5:52
  • $\begingroup$ Yeah. Thanks for help $\endgroup$ – Taylor Ted Dec 31 '15 at 5:53
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HINT: The equation can be written as:

$$y^3\frac{dy}{dx}=\frac{1}{x(1+xy^4)}$$

Put $y^4=t$

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