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As I am not very familiarized whit differential equations (I'm more from algebra), I don't know how to solve this problem, but I need to understand how it's done because I have to explain it to a friend. I hope anyone could help me.

Calculate the general solution of the next differential equation: $$yF(xy)dx=xG(xy)dy $$

By other side, I want to ask too where can I find a book to understand how to do this kind of problems.

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There is the following "well-known" differential equation:

$$yF(xy)dx+xG(xy)dy=0$$

This is very similar to the differential equation that you have provided in that we have instead $yF(xy)dx-xG(xy)dy=0$.

Whenever $F\neq G$ we have the integrating factor

$$\mu=\dfrac{1}{xy(F(xy)-G(xy))}$$

With the integrating factor, we arrive at an implicit solution, with a substitution $u=xy$:

$$\ln(x)=\int\dfrac{G(u)du}{u(G(u)-F(u))}+C$$

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Any introductory text on differential equations should have the basics. Of course, you will be expected to have some experience with Algebra and Calculus so that looking at a problem with "F(xy)" and "G(xy)" you immediately think "hmm, looks like the substitution u= xy would be worth trying". If u= xy then $y= u/x so dy= du/x- udx/x^2$. The equation $yF(xy)dx= xG(xy)dy$ becomes $(u/x)F(u)dx= xG(u)(du/x- udx/x^2)$. That is the same as $uF(u)du= xG(u)du- uG(u)dx$ so $(uF(u)- xG(u))du= -uG(u)dx$.

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  • $\begingroup$ No, you should have $uFdx$ not $uFdu$ $\endgroup$ – almagest Jun 21 '16 at 19:17

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