# Finding an Integral Surface

Consider finding the integral surface of

$$x^2 p + xy q = xyz-2y^2$$

which passes through the line $$x=y e^y$$ in the $$z=0$$ plane.

Attempt

In Lagrange's subsidiary form $$\frac{dx}{x^2}=\frac{dy}{xy}=\frac{dz}{(xyz-2y^2)}$$ Firstly consider $$\frac{dx}{x^2}=\frac{dy}{xy}$$ One can trivially show that $$a = \frac{x}{y}$$ where $$a$$ is an arbitrary constant. Now, consider $$\frac{dx}{x^2}=\frac{dz}{(xyz-2y^2)}$$ which may be written as $$\frac{dz}{dx}=\frac{(xyz-2y^2)}{x^2} \equiv \frac{z}{a}-\frac{2}{a^2}$$ having used $$a=x/y$$ from before. (After this point I am unsure of my working...) $$\frac{dz}{dx}=\frac{az-2}{a^2} \implies\frac{dz}{(az-2)}=\frac{dx}{a^2}$$

1. As $$a$$ is a function of $$a(x,y)$$, albeit an arbitrary constant, is my solution above sensical or have a made a mistake?

2. I understand that to find the integral surface the general solution is of the form $$F(a,b)$$ where has so far been determined to be $$a=x/y$$. How can I find this over arbitrary constant $$b$$?

Your calculus is correct. Don't worry about $$a(x,y)$$ which should be true outside the characteristic curves, but is constant on the characteristic curves. Thus it is legitim to integrate $$\frac{dz}{dx}=\frac{z}{a}-\frac{2}{a^2}$$ with $$a=$$ constant. $$z=\frac{2}{a}+be^{x/a}$$ The second family of characteristic curves is : $$e^{-x/a}(z-\frac{2}{a})=b$$ General solution of the PDE on the form of implicit equation $$F(a,b)=0$$ with any function $$F$$ of two variables, or equivalently $$b=\Phi(a)$$ with any function $$\phi$$ of one variable : $$b=e^{-x/a}(z-\frac{2}{a})=\Phi(\frac{x}{y})=e^{-y}(z-\frac{2y}{x})$$ $$z(x,y)=\frac{2y}{x}+e^y \Phi(\frac{x}{y})$$ $$\Phi$$ is an arbitrary function to be determined according to the boundary condition.
CONDITION : $$x=ye^y$$ on the plane $$z=0$$ . $$\quad z(ye^y,y)=0=\frac{2y}{ye^y}+e^y \Phi(\frac{ye^y}{y})$$ $$\Phi(e^y)=-2 e^{-2y}=-2(e^{y})^{-2}$$ So, the function $$\Phi$$ is determined : $$\Phi(X)=-2X^{-2}$$ . We put it into the above general solution where $$X=\frac{x}{y}$$ $$z(x,y)=\frac{2y}{x}-2e^y \frac{y^2}{x^2}$$
Carry out the second integration, $$az-2=be^{x/a}=be^y.$$ Then use $$b=\phi(a)$$ or a similar dependence and insert the initial condition $$-2 = \phi(e^y)e^{y}\implies \phi(t)=-\frac{2}t.$$