Question : Solve quasi linear pde x(y^2-z^2 )p+y(z^2-x^2 )q=z(x^2-y^2 )
Solution :
Given quasi linear pde
\[x\left(y^2-z^2\right)p+y\left(z^2-x^2\right)q=z\left(x^2-y^2\right)\]Comparing from standard equation of quasi linear pde P(x,y,z)p+Q(x,y,z)q=R(x,y,z)
Where p, q are partial derivative of z with respect to x and y respectively.
\[=>P=x\left(y^2-z^2\right)\ ,\ Q=y\left(z^2-x^2\right),\ R=z\left(x^2-y^2\right)\]Now solution of the pde is given as
\[\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\]\[=>\frac{dx}{x\left(y^2-z^2\right)}=\frac{dy}{y\left(z^2-x^2\right)}=\frac{dz}{z\left(x^2-y^2\right)}\]For the first solution Choosing multiplers x , y ,z then
\[\frac{dx}{x\left(y^2-z^2\right)}=\frac{dy}{y\left(z^2-x^2\right)}=\frac{dz}{z\left(x^2-y^2\right)}=\frac{xdx+ydy+zdz}{x^2\left(y^2-z^2\right)+y^2\left(z^2-x^2\right)+z^2\left(x^2-y^2\right)}\]\[\frac{dx}{x\left(y^2-z^2\right)}=\frac{dy}{y\left(z^2-x^2\right)}=\frac{dz}{z\left(x^2-y^2\right)}=\frac{xdx+ydy+zdz}{0}\]\[=>xdx+ydy+zdz=0\]
\[x^2+y^2+z^2=c\]To get the second solution lets choose multiplers 1/x , 1/y ,1/z then
\[\frac{dx}{x\left(y^2-z^2\right)}=\frac{dy}{y\left(z^2-x^2\right)}=\frac{dz}{z\left(x^2-y^2\right)}=\frac{\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz}{y^2-z^2+z^2-x^2+x^2-y^2}\]\[\frac{dx}{x\left(y^2-z^2\right)}=\frac{dy}{y\left(z^2-x^2\right)}=\frac{dz}{z\left(x^2-y^2\right)}=\frac{\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz}{0}\]\[=>\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz=0\]
\[=>\ log(x)\ +log(y)+log(z)\ =log(c)\]\[=>log(xyz)=log(c)\]\[=>\ xyz=c\]This is the second solution
Now general solution of the given pde is
\[z=f(x^2+y^2+z^2,\ xyz)\]where f is an arbitrary function.