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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 $b$?

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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 $X=\frac{x}{y}$ $$z(x,y)=\frac{2y}{x}-2e^y \frac{y^2}{x^2}$$

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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 $$ -2 = \phi(e^y)e^{y}\implies \phi(t)=-\frac{2}t. $$

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