# Understanding a step in solving a homogeneous differential equation

In an example from a book, a DE is $(x^2 + y^2)dx + (x^2 - xy)dy = 0$. It is solved by using the substitution

\begin{bmatrix} y = ux \\[0.3em] dy = x \; du + u \; dx \end{bmatrix}

So the equation becomes: $$(x^2 + u^2x^2)dx + (x^2 - ux^2)[u \; dx + x \; du] = 0$$

$$x^2(1 + u)dx + x^3(1 - u) du = 0$$

How do they get to this step, for the $x^3(1 - u) du$ part?

When I worked it out, I got $x^2[1 - u][x \; du + u \; dx]$, but I couldn't see how to proceed.

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Just algebra.

$$(x^2 + u^2x^2)dx + (x^2 - ux^2)[u dx + x du] = 0$$

Combining like terms, we arrive at:

$$(x^2 + u^2x^2 - u^2x^2 + ux^2)dx + (x^2 - u x^2)(x du) = 0$$

This reduces to:

$$x^2(1 + u)dx + x^3(1 - u) du = 0$$

Regards

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Oh! I didn't know you could distribute everything. – badjr Feb 8 '13 at 16:26
+1 nice done... – Babak S. Feb 8 '13 at 16:30
@BabakSorouh: thank you! Regards – Amzoti Feb 8 '13 at 16:33
Excellent: break it down to reveal its simplicity! +1 – amWhy May 3 '13 at 2:22
@amWhy: Thank you my friend! – Amzoti May 3 '13 at 2:27