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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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  • $\begingroup$ How do you get $(x-ux)dx+x(udx+xdu)=0$? I think this equation is wrong. $\endgroup$
    – PAD
    Sep 30 '12 at 22:25
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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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  • $\begingroup$ You're right! Thanks! $\endgroup$ Oct 3 '12 at 13:51
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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.

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