M6- Please I want solution of this question with all steps. Thanks a?udu6) Solve the partial differential equation= sin(x + y) given that at y = 0,дх ду1andaxx = 0,u = (y – 1)2. Explain in your words the impact of changing the given conditions toдиx = 1,= 7 and y = 0, u = (x – 1)2 on the final answer.ду

Given the partial differential equation is ∂2u∂x∂y=sinx+ygiven that at y=0, ∂u∂x=1 and x=0,…

Answered: a?u au = sin(x + y) given that at y =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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M6- Please I want solution of this question with all steps. Thanks

a?u
du
6) Solve the partial differential equation
= sin(x + y) given that at y = 0,
дх ду
1
and
ax
x = 0,
u = (y – 1)2. Explain in your words the impact of changing the given conditions to
ди
x = 1,
= 7 and y = 0, u = (x – 1)2 on the final answer.
ду

Expert Solution

Step 1

Given the partial differential equation is

$\frac{\partial^{2} u}{\partial x \partial y} = \sin (x + y) g i v e n t h a t a t y = 0, \frac{\partial u}{\partial x} = 1 a n d x = 0, u = {(y - 1)}^{2} a n d a t x = 1, \frac{\partial u}{\partial y} = 7 a n d y = 0, u = {(x - 1)}^{2}$

Step 2

Now

$\frac{\partial^{2} u}{\partial x \partial y} = \sin (x + y) a t y = 0, \frac{\partial u}{\partial x} = 1 a n d x = 0, u = {(y - 1)}^{2} \frac{\partial^{2} u}{\partial x \partial y} = \sin (x + y) . . . (1) I n t e g r a t e e q u a t i o n (1) w i t h r e s p e c t t o y, w e h a v e \frac{\partial u}{\partial x} = - \cos (x + y) + A . . . (2) p u t y = 0, \frac{\partial u}{\partial x} = 1 1 = - \cos (x + 0) + A A = \cos x p u t A = \cos x i n e q u a t i o n (2), w e h a v e \frac{\partial u}{\partial x} = - \cos (x + y) + \cos x . . . (3) I n t e g r a t e e q u a t i o n (3) w i t h r e s p e c t t o x, w e h a v e u = - \sin (x + y) + \sin x + B . . . (4) p u t x = 0, u = {(y - 1)}^{2} {(y - 1)}^{2} = - \sin (0 + y) + \sin 0 + B B = {(y - 1)}^{2} + \sin y p u t B = {(y - 1)}^{2} + \sin y i n e q u a t i o n (4), w e h a v e u = - \sin (x + y) + \sin x + {(y - 1)}^{2} + \sin y$

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