# General Solution to Quasilinear PDE using Method of Characteristics

This is a homework that I'm having a bit of trouble with. I posted it previously but there was a typo in my original post. Since I received an answer for the incorrect problem it was suggested that I post the correct problem as a new question:

Find a general solution of:

$(x^2+3y^2+3u^2)u_x-2xyu_y+2xu=0~.$

Of course this is to be done using the method of characteristics. The characteristic equations are:

$\dfrac{dx}{x^2+3y^2+3z^2}=-\dfrac{dy}{2xy}=-\dfrac{dz}{2xz}$

The idea is to find two functions, say $\phi$ and $\psi$, that are independent (the gradients are not colinear) and that are constant along the characteristics. We can then express the general solution as F($\phi$,$\psi$)=0 where F is an arbitrary $C^1$ function. It is easy to find that $\psi=\dfrac{y}{z}=const$. Finding a suitable $\phi$ is what I'm having trouble with. I tried substituting $\dfrac{y}{const}$ for $z$ in the $dx$ equation and solving for $\dfrac{dx}{dy}$ but the resulting ODE has no obvious (to me) method of solution.

Any suggestions?

• This is not an exact duplicate; the other post has extra factor $u$ in the second term. (My understanding is that this factor was a typo, but the other question was answered in this form.) – user53153 Jan 24 '13 at 17:29

Ok, I figured this out. It is clear that $y/z=\mathrm{const}=\psi$ will work. For $\phi$ we can substitute $z=y/c$. We end up with an exact equation:
$$2xy dx + (x^2+3y^2+\dfrac{3}{c^2}y^2) dy = \dfrac{d}{dy}(\phi(x(y),y)) = 0$$
with $$\phi=x^2y+y^3+\dfrac{yy^2}{c^2} = x^y+y^3+z^2y.$$
Clearly $\phi$ is a constant since its derivative is 0.
Thus: $F\left(\dfrac{u}{y},x^2y+y^3+u^2y\right)=0$ is the general solution for an arbitrary $F$ with continuous first derivatives. Using the implicit function theorem this is the same as:
$$u=yf(x^2y+y^3+u^2y).$$