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Question Number 124634 by bemath last updated on 05/Dec/20

S o l v e t h e e q u a t i o n \frac{\partial^{2} z}{\partial x \partial y} = x^{2} y

(1) p a r t i c u l a r s o l u t i o n w h i c h

z (x, 0) = x^{2} a n d z (1, y) = \cos y

Answered by liberty last updated on 05/Dec/20

(∙) \frac{\partial}{\partial x} (\frac{\partial z}{\partial y}) = x^{2} y . I n t e g r a t i n g w i t h r e s p e c t t o x

w e f i n d \frac{\partial z}{\partial y} = \frac{1}{3} x^{3} y + F (y); F (y) i s a r b i t r a r y

i n t e g r a t i n g a g a i n w i t h r e s p e c t t o y

\Rightarrow z = \frac{1}{6} x^{3} y^{2} + H (y) + G (x)

w h e r e G (x) i s a r b i t r a r y

s i n c e z (x, 0) = x^{2} w e h a v e x^{2} = H (0) + G (x)

o r G (x) = x^{2} - H (0)

t h u s z = \frac{1}{6} x^{3} y^{2} + H (y) + x^{2} - H (0)

s i n c e z (1, y) = \cos y w e h a v e

\cos y = \frac{1}{6} y^{2} + H (y) + 1 - H (0) o r

H (y) = \cos y - \frac{1}{6} y^{2} - 1 + H (0)

t h e r e f o r e z = \frac{1}{6} x^{3} y^{2} + \cos y - \frac{1}{6} y^{2} - 1 + H (0) + x^{2} - H (0)

w e g e t t h e r e q u i r e d s o l u t i o n

z = \frac{1}{6} x^{3} y^{2} + \cos y - \frac{1}{6} y^{2} + x^{2} - 1 .

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