# Does Charpit's method gives general solution to first order non linear partial differential equations? What about this counter example?

Using Charpit's method, the solution to the PDE

$\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=0$

comes out in the form:

$z=ax-ay+c$, as claimed by many textbooks.

But consider this alternate unaccounted solution to the same problem:

$z=\frac{x^2}{2}-xy+\frac{y^2}{2}$

Similar approach can yield a class of different solutions in the form of polynomials of higher degree in $x$ and $y$.

Charpit's method does not yields these solutions. (Does it?). So, how can we call it a general method to solve first order PDEs?

$z=ax-ay+c$ isn't the general solution.
The general solution, thanks to Charpit's method or other methods is : $$z=F(x-y)$$ where $F(X)$ is any differentiable function.
For example with $F(X)=aX+c \quad\to\quad z=a(x-y)+c=ax-ay+c$
Other example, with $F(X)=\frac{1}{2}X^2 \quad\to\quad z=\frac{1}{2}(x-y)^2 = \frac{x^2}{2}-xy+\frac{y^2}{2}$
You can obtain an infinity of solutions of the PDE with other functions $F(X)$