Answer to Question #149879 in Differential Equations for Nikhil

Question

Solve the simultaneous equation
dx/(x-y)y^2 = dy/-(x-y)x^2 = dz/z(x^2+y^2)

Answer 1

Given equations are,

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

Taking first two,

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

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

Integrating both sides , we get

$\int -x^2dx = \int y^2dy$

$y^3 = -x^3 + k$ (1)

then, $y = (k-x^3 )^{1/3}$

Taking first and last part,

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

Putting value of y, we get,

$\frac{(x^2+y^2)}{y^2x-y^3}dx = \frac{1}{z}dz$

$\frac{(x^2+(k-x^3)^{2/3})}{(k-x^3)^{2/3}x-(k-x^3)}dx = \frac{1}{z}dz$

Now integrating both sides, we get

$\frac{1}{6} (-\frac{(3 x^2 (\frac{k - x^3}{k - 2 x^3})^{(2/3)} 2^{F1(\frac{2}{3}, \frac{2}{3}; \frac{5}{3}; -\frac{x^3}{(k - 2 x^3)}}} {(k - x^3)^(2/3)} ) + 2 log(2 x^3 - k) + 2 log(1 - \frac{x}{(k x^3 - 1)^{(1/3)}}) - 2 \sqrt{3} tan^{-1}(\frac{\frac{2 x}{(k x^3 - 1)^{1/3}} + 1}{\sqrt{3}}) - log(\frac{x}{(k x^3 - 1)^{1/3} }+ \frac{x^2}{(k x^3 - 1)^{2/3}} + 1)) + C = lnz$

Where $2^{F_1(a,b;c;x )}$ is the hypergeometric function

*Integration is done by online calculator as solution by hand is not possible for the differential equation.