Substitution with $a = x$, $b=\frac{y-x^2}{x} $ in differential equation gives me $ u = a \frac{\partial u}{\partial a} $ - Why?

Question

If

$$ \cases{a = x\\ b=\frac{y-x^2}{x}} $$

and

$$ x\frac{\partial u}{\partial x}+\left(y+x^2\right)\frac{\partial u}{\partial y}=u $$

Why do we get

$$ u = a \frac{\partial u}{\partial a} $$

Answer 1

HINTS:

Use the chain rule see this example

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial u}{\partial b}\frac{\partial b}{\partial x}$$

and

$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial a}\frac{\partial a}{\partial y}+\frac{\partial u}{\partial b}\frac{\partial b}{\partial y}$$

Note that $x=a$, $y=ab+a^2$, $\frac{\partial a}{\partial x}=1$, $\frac{\partial a}{\partial y}=0$, $\frac{\partial b}{\partial y}=-2-b/a$, and $\frac{\partial b}{\partial x}=1/a$.