# Thread: partial differential equation satisfied by u(x,y)

1. ## partial differential equation satisfied by u(x,y)

Let f : R $\rightarrow$ R be differentiable and define u: $R^2$ $\rightarrow$ $R$
by u(x,y) := $e^{x sin y}f(x - y)$ show that u satisfies the partial differential equation
$du/dx$ + $du/dy$= (sin y + x cos y)u ?

2. $\displaystyle{\frac{du}{dx} = \sin y\ e^{x \sin y} f(x-y) + e^{x \sin y} f'(x-y) \frac{df}{dx}}$

Similarly for $\frac{du}{dy}$, then add the two, obviously... pic in a mo.

... where (key in spoiler) ...

Spoiler:

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to either x or y.

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