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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

\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 .