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$$ \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 , u(x,0) = \sin x , u(0,y) = y $$

I've tried to solve it and this what I do

$$ \frac{\partial}{\partial x } \left( \frac{\partial u}{\partial y } \left( x,y \right) \right)= 0 $$

integrate with respect to $x$ yields

$$\frac{\partial u}{\partial y } \left( x,y \right)= f(y) $$

integrate with respect to $y$ yields

$$ u( x,y )= F(y) + g(x) $$

with $$ \frac{\partial }{\partial y } F(y) = f(y) $$

is this correct?

what's should I do next?

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Don't write with that awful \large command. Better, use double dollar signs to center and increase size of font. – DonAntonio Feb 11 '13 at 13:59
hmm..., ok..., thanks for revising.. :) – chihiroasleaf Feb 11 '13 at 14:19

What you should do next is use your initial conditions:

$$u(x,0) = \sin{x} \implies g(x) + F(0) = \sin{x}$$

$$u(0,y) = y \implies g(0) + F(y) = y$$

$$u(x,y) = F(y) + g(x) = y - g(0) + \sin{x} - F(0) = y + \sin{x} + C$$

where $C$ is some constant. We know the value of this constant because we also know that $u(0,0) = 0$ (plug zero into either initial condition); thus, $C=0$ and

$$u(x,y) = y + \sin{x}$$

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where is $g(0) and F(0)$ ? are they become $C$ ? – chihiroasleaf Feb 11 '13 at 19:41
$C=-F(0)-g(0)$; the constants may be combined into a single constant. – Ron Gordon Feb 11 '13 at 19:46
ahh.., I see.. :) and are we allowed to add the 'initial condition' to find C ? – chihiroasleaf Feb 11 '13 at 20:09
Yes...that's what my solution does. I use the conditions to find $C$, and thus $u$, uniquely. – Ron Gordon Feb 11 '13 at 20:14

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