# quasi linear pde

I have a pde of the form: $$x(y^2-u^2)u_x+y(u^2-x^2)u_y=u(x^2-y^2)$$ and I need to find general solution

the method of characteristics gives system of ode that I cannot solve

all I found is that $$x^2+y^2+u^2$$ is constant on every characteristic curve

Is there any trick I am missing?

thank you

• what do you mean? I differentiated the expression $x^2+y^2+u^2$ according $t$ and used the relations: $\dot{x}=a=x(y^2-u^2) , \dot{y}=b=y(u^2-x^2), \dot{u}=c$ and got it is 0 – lisa Oct 28 '15 at 12:02

I agree with $(x^2+y^2+u^2)$= constant for the first.
The second is $(x\;y\;u)$=constant, leading to the general solution :
The result can be expressed on various equivalent forms. For example : $$x\,y\,u=f(u^2+x^2+y^2)$$ with any derivable function $f$.