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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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1 Answer 1

up vote 3 down vote accepted

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... –  B. 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

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