Answer to Question #146748 in Differential Equations for Nikhil Singh

Question

Solve the simultaneous differential equation
dx/x^2-y^2-z^2=dy/2xy=dz/2xz

Answer 1

Given differential equation is, $\frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$

Taking last two terms, $\frac{dy}{2xy}=\frac{dz}{2xz} \implies \frac{dy}{y}=\frac{dz}{z}$

Integrating both sides, we get,

$y=cz$ (1)

Taking first and last term,

$\frac{dx}{x^2-y^2-z^2}=\frac{dz}{2xz}$

$\frac{dx}{x^2-c^2z^2-z^2}=\frac{dz}{2xz}$

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

Putting x=vz, then, $\frac{dx}{dz} = v+z\frac{dv}{dz}$

Then equation will be, $v+z\frac{dv}{dz} = \frac{v^2x^2-z^2(1+c^2)}{2vz^2} = \frac{v^2-(1+c^2)}{2v}$

$z\frac{dv}{dz} = -\frac{v^2+(1+c^2)}{2v}$

$\frac{2vdv}{v^2+(1+c^2)} = -\frac{dz}{z}$

Integrating the above equation, $log|v^2+(1+c^2)|=-log|z|+log|c_1| \implies v^2+(1+c^2)=\frac{c_1}{z}$

Putting value of v, $\frac{x^2}{z^2}= \frac{c_1}{z} -(1+c^2)$

$x^2 = c_1z - z^2(1+c^2)$

Solving quadratic equation for z, we get,

$z=\frac{c_1\pm \sqrt{(c_1^2-4x^2(1+c^2)})}{ 2(1+c^2)}$ (2)

Equation (1) & (2) are solutions.