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Solving Partial Differential Equation $u_{xyy} (x,y,z) = 2 \sin x $


Is this correct?

$$ \frac{\partial^3 u}{\partial y \, \partial y \, \partial x} (x,y,z) = 2 \sin x $$

Integrate respect to y

$$ \frac{\partial^2 u}{ \partial y \, \partial x} (x,y,z) = 2y \sin x + f(x) $$

Integrate respect to y

$$ \frac{\partial u}{ \partial x} (x,y,z) = y^2 \sin x + yf(x) + g(x) $$

Integrate respect to x $$ u(x,y,z) = -y^2 \cos x + yF(x) + G(x) + h(y) ;\quad \frac{\partial}{\partial x}F(x) = f(x),\quad \frac{\partial}{\partial x} G(x) = g(x) $$

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3  
Why aren't $f$ and $g$ functions of both $x$ and $z$? –  Ayman Hourieh Feb 13 '13 at 0:42
    
so.., after the first integration, it should be $f(x,z)$ and in second integration $g(x,z)$, right? *I'm a bit confused integrating in this form.. * and in the last integration, is it correct $h(y)$? or it should be $h(y,z)$ –  chihiroasleaf Feb 13 '13 at 0:47
    
Yes. The last integration should add a function of $y, z$. –  Ayman Hourieh Feb 13 '13 at 0:49
    
wait.., if it is $f(x,z)$ then what's the result if I integrate $yf(x,z)$ respect to $x$ ? –  chihiroasleaf Feb 13 '13 at 0:50
    
Integrating $y f(x,z)$ with respect to $x$ will give you $y F(x,z) + G(y,z)$ for some functions $F$ and $G$, where $\partial f/\partial x = F$. –  Robert Israel Feb 13 '13 at 1:08

1 Answer 1

$$ u_{xyy} = 2\sin x \\ u_{xy} = 2y\sin x + f(x,z)\\ u_x = y^2 \sin x + yf(x,z) + g(x,z) \\ u = -y^2 \cos x + yF(x,z) + G(x,z) + H(y,z) $$ where $F$ and $G$ are completely arbitrary functions ($F \in C^1$).

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at the second integration, why there is no constant if integration? isn't should be $ \frac{\partial u}{ \partial x} (x,y,z) = y^2 \sin x + yf(x) + g(x,z) $ –  chihiroasleaf Feb 13 '13 at 7:58
    
It should, of course. My bad –  Kaster Feb 13 '13 at 8:00

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