Answer to Question #113582 in Differential Equations for Raghavaraju Nithisha

Question

Solve : x(y^2-z^2)p+y(z^2-x^2)q=z(x^2-y^2)

Answer 1

The given equation is a Lagrange's linear equation. The auxiliary equations are

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

Using the multipliers, $x,y,z$ each of the equation is equal to

$\dfrac{xdx+ydy+zdz}{x^{2}(y^{2}-z^{2})+y^{2}(z^{2}-x^{2})+z^{2}(x^{2}-y^{2})}$

Therefore,

$xdx+ydy+zdz=0\\ \text{Integrating, we get}\\ \dfrac{x^{2}}{2}+\dfrac{y^{2}}{2}+\dfrac{z^{2}}{2}=c\\ x^{2}+y^{2}+z^{2}=c_{1}~~~~~~~~~~~~(1)$

Now, using the multipliers, $\dfrac{1}{x}, \dfrac{1}{y}, \dfrac{1}{z}$ each of the equation is equal to

$\dfrac{\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz}{y^{2}-z^{2}+z^{2}-x^{2}+x^{2}-y^{2}}$

Therefore,

$\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz=0\\ \text{Integrating, we get}\\ \log x+\log y+\log z = \log c_{2}\\ \log (xyz)=\log c_{2}\\ xyz = c_{2}~~~~~~~~~~~~~~~~~~~~~~~~(2)$

The general solution is given by,

$\phi(c_{1},c_{2})=0\\ \phi(x^{2}+y^{2}+z^{2},xyz)=0\\$