Answer to Question #188636 in Differential Equations for mahammad

Question

erify that the following cquations are integrable and determine their solutions.
x(y^2-a^2 )dx+y(x^2-t^2 )dy-z(y^2-α^2 )dz=0, where α is constant

Answer 1

The given equation is-

$x(y^2-a^2 )dx+y(x^2-t^2 )dy-z(y^2-α^2 )dz=0,$

Given equation is of the form-

$Pp+Qq+Rr=0$

The $\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}$

So,$\dfrac{dx}{x(y^2-a^2)}=\dfrac{dy}{y(x^2-t^2)}=\dfrac{dz}{z(y^2-\alpha^2)}$

Taking first two terms-

 $\dfrac{dx}{x(y^2-a^2)}=\dfrac{dy}{y(x^2-t^2)}$

$\Rightarrow \dfrac{(x^2-t^2)dx}{x}=\dfrac{(y^2-a^2)dy}{y}$

Integrating both the sides-

$\dfrac{x^2}{2}-t^2lnx=\dfrac{y^2}{2}-a^2lny+c_1$

$\Rightarrow \dfrac{x^2}{2}-t^2lnx-\dfrac{y^2}{2}+a^2lny=c_1~~~~~~~~~-(1)$

Taing last two terms-

$\dfrac{dy}{y(x^2-t^2)}=\dfrac{dz}{z(y^2-\alpha^2)}$

$\Rightarrow \dfrac{(y^2-\alpha^2)dy}{y}=\dfrac{(z^2-t^2)dz}{z}$

Integrating Both the sides-

$\dfrac{y^2}{2}-\alpha^2lny=\dfrac{z^2}{2}-t^2lnz+c_2$

$\Rightarrow \dfrac{y^2}{2}-\alpha^2lny-\dfrac{z^2}{2}+t^2lnz=c_2$

Hence The solution is-

$\phi(c_1,c_2)=0$

$\phi (\dfrac{x^2}{2}-t^2lnx-\dfrac{y^2}{2}+a^2lny, \dfrac{y^2}{2}-\alpha^2lny-\dfrac{z^2}{2}+t^2lnz)=0$