Why does the "solution" not satisfy the PDE?

Question

I'm trying to solve the following PDE:

$$x z_x -xyz_y = z \quad with\quad z(x,x) =x^2 e^x \quad (x,y) \in \mathbb{R}^2.$$

I've proceeded as, $$\frac{dx}{ x} = \frac{dy }{-xy } \Rightarrow y = C e^{-x}, $$ so $$\frac{dz}{ dx} = z \Rightarrow z = G(C) e^x .$$

Using the given curve, $$ \begin{split} z(x,x) &= G\left(\frac{ x}{ e^{-x}}\right) e^x = x^2 e^x\\ \\ & \Rightarrow G(t) = t^2 e^{2x} (\text{ I am not sure this is the only possible form of }G). \end{split} $$ Hence,

$$z(x,y) = y^2 e^x,$$ but the thing is, this does not satisfy the given PDE: $$ x (y^2 e^x) - xy (2y e^x) = -xy^2e^x \not = z. $$ I'm assuming that there is a problem while finding the characteristic curve since I cancelled $x$s there, but not sure why would that cause a problem, so I'm looking for both a solution and an explanation what is wrong in my wrong solution.

Answer 1

You need to consider the full set of Lagrange equations $$ \frac{dx}{x}=\frac{dy}{-xy}=\frac{dz}{z} $$ so that indeed $ye^x=c_1$, but from the first and third equation you get $\ln x = \ln z +C\implies z=c_2x$, so that in conclusion the general solution form is $$ z=x\phi(ye^x) $$

Answer 2

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=-xy=-e^ty$ , letting $y(0)=y_0$ , we have $y=y_0e^{1-e^t}=y_0e^{1-x}$

$\dfrac{dz}{dt}=z$ , letting $z(0)=f(y_0)$ , we have $z(x,y)=f(y_0)e^t=f(e^{x-1}y)x=F(e^xy)x$

$z(x,x)=x^2e^x$ :

$F(xe^x)x=x^2e^x$

$F(xe^x)=xe^x$

$F(x)=x$

$\therefore z(x,y)=xye^x$