Show that a partial differential eqution is satisfied (is my notation okay?)

Let:

$f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable.

$u:\mathbb{R}^2\rightarrow\mathbb{R}$ and $u(x,y)=e^{-y}f(x+y^2)$

Show that: $$2y\frac{\partial{u}}{\partial{x}}-\frac{\partial{u}}{\partial{y}}=u$$

My attempt has been to first work out the two partial derivatives and plug them in, however I am not very confident in my notation, for example:

$e^{-y}\frac{\partial{f(x+y^2)}}{\partial{y}}$ is the second part of my working of partial u/partial y, first * derivative of second

The next step is: $2ye^{-y}\frac{\partial{f(x+y^2)}}{\partial{y}}$ but that looks the same as before, my notation doesn't show f(x+y^2) and the partial derivative of f wrt a variable at a point as being different. How do I overcome this?

I've never been formally taught or had the need for the vertical bar notation, but I sense this is what the notation solves, how do I write it?

Addendum: I've not LaTeXed a lot of it, I find it a laggy choir at present, please don't take this as laziness, I'm also going to try introducing a new variable used for x+y^2 but on paper. If I answer it I shall report back, sorry for mixing up my mu and u, it's one of those prints of a scan of an old text-book, the toner is shiny on the cheap paper and it's black and white, not gray-scale a choir to read!

Addendum 2: Here, it works, but I'm not sure if it's errors canceling or not:

Let: $\lambda:\mathbb{R}^2\rightarrow\mathbb{R}$, $\lambda(x,y)=x+y^2$

Now:

$u(x,y) = e^{-y}f(\lambda)$ (I feel I need not write $\lambda(x,y)$ all the time)

$$\frac{\partial{u}}{\partial{x}}=e^{-y}\frac{df(\lambda)}{\partial{x}}=e^{-y}\frac{df}{d\lambda}.\frac{\partial{\lambda}}{\partial{x}}=e^{-y}\frac{df}{d\lambda}$$

and

$$\frac{\partial{u}}{\partial{y}}=e^{-y}\frac{df(\lambda)}{\partial{y}}+\frac{\partial}{\partial{y}}[e^{-y}]f(\lambda)$$ Now there I've written partial by partial-y, this is why I am not that happy with my notation.... but d/partial y doesn't look right either...

Anyway: $$\frac{\partial{u}}{\partial{y}}=-e^{-y}f(\lambda)+e^{-y}\frac{df}{d\lambda}.\frac{\partial{\lambda}}{\partial{y}}=e^{-y}[2y\frac{df}{d\lambda}-f(\lambda)]$$

To show that is just to bung those in.

• Did you misplace $u$ instead of $\mu$? – Kaster Aug 29 '13 at 19:56
• See addendum, I'll change it though as someone agrees with your comment – Alec Teal Aug 29 '13 at 19:59
• @Kaster is $\frac{\partial{f(\lambda)}}{\partial{x}}=\frac{\partial{f}}{\partial{\lambda}}.\frac{\partial{\lambda}}{\partial{x}}$ allowed? Where $\lambda:\mathbb{R}^2\rightarrow\mathbb{R}$ and $\lambda(x,y)=x+y^2$ or is it a matter of errors canceling. It seems logical at least. But I'd prefer to write $\frac{df}{d\lambda}$ – Alec Teal Aug 29 '13 at 20:14

$$\lambda (x,y) = x+y^2 \\ u = e^{-y}f(\lambda) \\ u_x = e^{-y} f'(\lambda) \lambda_x = e^{-y} f'(\lambda)\\ u_y = -e^{-y}f(\lambda) + e^{-y} f'(\lambda) \lambda_y = e^{-y} \left[2yf'(\lambda) -f(\lambda)\right ] \\ 2y u_x - u_y = 2y e^{-y} f'(\lambda) - 2ye^{-y}f'(\lambda) + e^{-y} f(\lambda) = e^{-y} f(\lambda) = u$$