I'm solving the PDE:
$$\exp(u)u_x + \frac{y}{x}u_y = 1$$
With characteristics given by
$$\frac{dx}{\exp(u)} = \frac{dy}{y/x} = \frac{du}{1} $$
Giving a corresponding solution of
$$c1 = x^2 -2y^2\exp(u)\\ c2 = u-\frac{y^2}{2x}$$
Thus
$$u(x,t) = \frac{y^2}{2x}+ F(x^2 - 2y\exp(u))$$ .
Is this right?
$$\exp(u)u_x + \frac{y}{x}u_y = 1$$ I agree with $\frac{dx}{\exp(u)} = \frac{dy}{y/x} = \frac{du}{1} $
First characteristic from solving $\frac{dx}{\exp(u)} = \frac{du}{1} $ $$e^u-x=c_1$$ Second characteristic from solving $\frac{dx}{\exp(u)} = \frac{dy}{y/x} \quad\implies\quad \frac{dx}{x+c_1} = x\frac{dy}{y}\quad\implies\quad \frac{dx}{x(x+c_1)} = \frac{dy}{y}$ $$\frac{x+c_1}{x}y^{c_1}=c_2$$ General solution on the form of implicit equation : $\frac{x+c_1}{x}y^{c_1}=F(c_1)$ $$\frac{e^u}{x}y^{e^u-x}=F(e^u-x)$$