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How to solve this ordinary differential equation: $$ (y^{3}+4e^{x}y)dx + (2e^{x}+3y^2)dy = 0 $$ This is from the book Fundamentals of Differential Equations by Nagle, Saff and Sniders. The eight edition on the page 77 problem 12.

I appreciate any hints.

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  • $\begingroup$ can you send us a link? $\endgroup$ – Dr. Sonnhard Graubner Sep 19 '16 at 10:55
  • $\begingroup$ the solution looks terrible $\endgroup$ – Dr. Sonnhard Graubner Sep 19 '16 at 11:16
  • $\begingroup$ What link should I send you? $\endgroup$ – Jure Vreča Sep 19 '16 at 13:46
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$$(y^{3}+4e^{x}y)dx + (2e^{x}+3y^2)dy = 0\to\\ (y^{3}+4e^{x}y)+ (2e^{x}+3y^2)y^{'} = 0$$ Consider substitution $y=te^{Ax}$, where $t-$function of $x$, $A-$some constant. $$ \left(t^3e^{3Ax}+4te^{(1+A)x}\right)+(2e^x+3t^2e^{2Ax})(t^{'}e^{Ax}+Ate^{Ax})=0\to\\ e^{3Ax}\left(t^3+4te^{(1-2A)x}\right)+e^{3Ax}(3t^2+2e^{(1-2A)x})(t^{'}+At)=0 $$ Notice that equation has the simplest form if we choose $A=\frac{1}{2}$. Also, as far as $A$ is finite, we can omit multiplicator $e^{3Ax}$. $$ \left(t^3+4t\right)+(3t^2+2)\left(t^{'}+\frac{t}{2}\right)=0\to t^{'}=-\frac{t(t^2+4)}{3t^2+2}-\frac{t}{2}\to\\ \frac{2(3t^2+2)}{5t(t^2+2)}dt=-dx $$ In terms of partial fractions: $$ \frac{2}{5}\left(\frac{1}{t}+\frac{2t}{t^2+2}\right)dt=-dx $$ Now it is easy to integrate $$ \frac{2}{5}\left(\ln(t)+\ln(t^2+2)\right)=-x+C\to\frac{2}{5}\ln(t(t^2+2))=C-x\to\\ (t^3+2t)^{\frac{2}{5}}=e^{C-x} $$ Back to the initial variables $t=ye^{-\frac{1}{2}x}$ $$ (y^3e^{-\frac{3}{2}x}+2ye^{-\frac{1}{2}x})^{\frac{2}{5}}=e^{C-x}\to\\ (y^3e^{-x}+2y)^{\frac{2}{5}}=e^{C-\frac{4x}{5}} $$

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Hint:

$(y^3+4e^xy)~dx+(2e^x+3y^2)~dy=0$

$(2e^x+3y^2)~dy=-(y^3+4e^xy)~dx$

$(3y^2+2e^x)\dfrac{dy}{dx}=-y^3-4e^xy$

Let $u=y^2$ ,

Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$

$\therefore\dfrac{3y^2+2e^x}{2y}\dfrac{du}{dx}=-y^3-4e^xy$

$(3y^2+2e^x)\dfrac{du}{dx}=-2y^4-8e^xy^2$

$(3u+2e^x)\dfrac{du}{dx}=-2u^2-8e^xu$

This belongs to an Abel equation of the second kind.

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  • $\begingroup$ I am not quite sure how to continue from this. But I was able to solve it another way, so I didn't pursue it to far. Thanks anyway. $\endgroup$ – Jure Vreča Sep 19 '16 at 15:24
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You need to find the intergration factor, which in this case comes out to be e^x . Multiply whole equation with it and you see that you have perfect differentials

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