Help with Change of Variables for a First Order PDE

Question

I am currently struggling with solving a PDE using the Changing of Variables Method. The equation is as follows:

$y^2U_x - xyU_y = xU - 2xy \ $

Now, I understand the basics of Changing of Variables, but I am struggling to find a value for S. My T value was found as $T = y^2-x^2$ by solving the equation $dx/y^2 = -dy/xy$ for the constant, which I then called T. As I understand it, to continue with Changing of Variables Method, I need to find a value for S, which is technically arbitrary, but surely there is a way to find one which makes the equation much easier without testing multiple solutions?

Thank you for your time, let me know if anything is unclear.

J

By $Ux$ do you mean $U_x$ ? There is a big difference when using or not subscript. — Integral, Jul 02 '17 at 20:39
Yes, sorry I was unsure how to do a subscript. For reference, it is the partial derivative of U with respect to x — J Sutherland, Jul 02 '17 at 22:48

score 1 · Answer 1 · answered Jul 03 '17 at 17:34

$$y^2U_x - xyU_y = xU - 2xy \tag 1$$ This is a non-homogeneous first order linear PDE.

An obvious particular solution is $U=y$. This draw us to the change of function : $$U(x,y)=V(x,y)+y$$ which changes the PDE (1) into (2) : $$y^2V_x - xyV_y = xV \tag 2$$ which is an homogeneous first order linear PDE, easy to solve thanks to the method of characteristics.

The set of ODE of the characteristic curves is : $\quad\frac{dx}{y^2}=\frac{dy}{-xy}=\frac{dV}{xV}$

A first family of characteristic equations comes from $\quad\frac{dx}{y^2}=\frac{dy}{-xy}\quad\to\quad x^2+y^2=c_1$

A second family of characteristic equations comes from $\frac{dy}{-xy}=\frac{dV}{xV}\quad\to\quad yV=c_2$

The general solution of the PDE (2) is : $\quad yV=f(x^2+y^2)\quad$ where $f(X)$ is any differentiable function with $X=x^2+y^2$.

$$V(x,y)=\frac{f(x^2+y^2)}{y}$$ $$U(x,y)=\frac{f(x^2+y^2)}{y}+y$$

Help with Change of Variables for a First Order PDE

1 Answers1