Answer to Question #179783 in Differential Equations for kavya

Question

solve dx/(x-y)y^2 = dy/(y-x)x^2 = dz/x^2+y^2

Answer 1

The question is

$\bigstar$

$dx\over { ((x-y)y^2) }$ =$dy\over((y-x)x^2)$ = $dz\over(x^2+y^2)$

Consider these

$dx\over { ((x-y)y^2) }$ =$dy\over((y-x)x^2)$

We get

Integration of

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$\boxed{\int y^2dy = \int -x^2dx}$

$\bull$

$\boxed{y^3 = -x^3+c1}$

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${dx\over ((x-y)y^2)}={dz\over(x^2+y^2)}$

 $\implies$

$\boxed{ \int {y^2} dz= \int {(x^2+y^2)\over (x-y) }dx }$

$\implies$

$\boxed{∫ {(x^2+y^2)\over (x-y) }dx}$ =$\boxed{∫ {2y^2\over(x-y)+x+y)}dx}$ =

$\boxed{2y^2\intop{1\over(x-y)}dx +\intop x dx + y \intop 1 dx}=$

= $\boxed{2 y^2 ln(x-y) + {x^2\over{2 }}+ xy + c_2}$

$\bull$

$\bigstar$

$\boxed{y^2ln(z) = 2 y^2 ln(x-y) + {x^2\over{2 }}+ xy + c_2 }$

integral surface:

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$\boxed{ y^3 = -x^3+c_1 }$

$\bull$

$\boxed{y^2ln(z) = 2 y^2 ln(x-y) + {x^2\over{2 }}+ xy + c_2 }$