Solve the equation: (∂^2 u/(∂x∂y)) = sin(x)cos(y), subjected to the boundary conditions at y = (π/2), (∂u/∂x) = 2x and x = π, u = 2sin(y)


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Question Number 23445 by tawa tawa last updated on 30/Oct/17

$Solve the equation : \frac{\partial^{2} u}{\partial x \partial y} = \sin (x) \cos (y), subjected to the boundary$ $conditions at y = \frac{π}{2}, \frac{\partial u}{\partial x} = 2 x and x = π, u = 2 \sin (y)$
	Answered by mrW1 last updated on 31/Oct/17
	$(∂^2 u/(∂x∂y)) = sin(x)cos(y) ⇒(∂u/∂x)=∫sin (x) cos (y) dy=sin (x) sin (y)+f(x) sin (x) sin ((π/2))+f(x)=2x ⇒f(x)=2x−sin (x) (∂u/∂x)=sin (x) [sin (y)−1]+2x ⇒u=∫{sin (x) [sin (y)−1]+2x}dx =−cos (x)[sin (y)−1]+x^2 +C −cos (π)[sin (y)−1]+π^2 +C=2 sin (y) ⇒C= sin (y)+1−π^2 ⇒u(x,y)=cos (x)+[1−cos (x)] sin (y)+x^2 +1−π^2$
	$\frac{\partial^{2} u}{\partial x \partial y} = \sin (x) \cos (y)$ $\Rightarrow \frac{\partial u}{\partial x} = \int \sin (x) \cos (y) dy = \sin (x) \sin (y) + f (x)$ $\sin (x) \sin (\frac{π}{2}) + f (x) = 2 x$ $\Rightarrow f (x) = 2 x - \sin (x)$ $\frac{\partial u}{\partial x} = \sin (x) [\sin (y) - 1] + 2 x$ $\Rightarrow u = \int {\sin (x) [\sin (y) - 1] + 2 x} dx$ $= - \cos (x) [\sin (y) - 1] + x^{2} + C$ $- \cos (π) [\sin (y) - 1] + π^{2} + C = 2 \sin (y)$ $\Rightarrow C = \sin (y) + 1 - π^{2}$ $\Rightarrow u (x, y) = \cos (x) + [1 - \cos (x)] \sin (y) + x^{2} + 1 - π^{2}$
	Commented by tawa tawa last updated on 31/Oct/17

	$God bless you sir .$
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