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An example in my Differential Equations textbook shows how to solve the homogenous differential equation $ (x^2+y^2)\,dx +(x^2-xy)\,dy=0 $

by substituting $y$ with $ux$, which I am trying to understand. The book explains that the reason we do this is so that $dy$ will equal $u\,dx + x\,du$.

The answer says that after substitution, the equation becomes $(x-ux)\,dx + x(u\,dx + x\,du) = 0 $ and then $ dx + x\,du = 0$

My question is, how did it get to $dx+x\,du =0$? Is it a typo or am I missing something?

I think it should be $ x\,dx + x\,du$ and then $dx + du$ and then $x + u$ and eventually $x + y/x$.

However, the textbook says the answer is $x\ln(x)+y=cx$.

What am I missing?

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    How do you get $(x-ux)dx+x(udx+xdu)=0$? I think this equation is wrong.2012-09-30

2 Answers 2

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$(x-ux)dx+x(udx+xdu)=0$ Expanding both sides gives $xdx-uxdx+xudx+x^2du=0$ $xdx+x^2du=0$ Assuming $x$ is non-zero: $dx+xdu=0$

You forgot the extra factor of $x$ attached to the $du$.

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    You're right! Thanks!2012-10-03
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Write $\frac{dy}{dx}=\frac{x^2+y^2}{xy-x^2}$

Then $ x \frac{du}{dx}+u=\frac{1+u^2}{u-1} $ This is a separable equation, which you can solve.