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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.