Help with PDE $xu_x+yu_y=2u$

Question

PDE looks like this: $$xu_x+yu_y=2u$$ $$u(x,1)=x^2$$ $u$ is a function of two variables $u(x,y)$ I'm new at this part of mathematics so I need help.
My first idea is to use transport PDE $u_t+cu_x=0$
(little change o variable $t$ with $y$) but the coefficients with $u_x$ isn't constant so I have another problem.

Answer 1

If you stare at this long enough, you'll see that $u = x^2$ is a solution.

(Or you can use the method of characteristics...)

Answer 2

Hint:

1. Oversve that this is an equation of the form $a(x,y,u)u_x+b(x,y,u)u_y=f(x,y,u)$

2. The solution to this equation can be found by comparing below ratios:

$$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{2u}$$

3. Find two linearly independent solutions $h(x,y,u)=c_1$ and $g(x,y,u)=c_2$ The general solution to given equation is given by $c_2=F(c_1)$

4. After that use initial condition to get final solution.

Answer 3

By chance is the initial condition $u(x,0) = x^2$? This is due to: \begin{align} \frac{du}{2 \, u} &= \frac{d(x + y)}{x+y} \\ \ln u &= 2 \, \ln(x + y) + \ln(c_{0}) \\ u(x,y) &= c_{0} \, (x+y)^{2}. \end{align}

Check: \begin{align} x \, u_{x} + y \, u_{y} &= x \, 2 c_{0} (x+y) + y \, c_{0} (x+y) \\ &= 2 \, c_{0} (x+y)^{2} \\ &= 2 u. \end{align}

If $u(x,0) = x^2$ then the solution is as is. If $u(x,1) = x^2$ then there is a potential issue.

Answer 4

The left-hand side $x u_x + y u_y$ is a directional derivative in the direction $(x,y)$, i.e., the radial direction, which suggests that problem will become easier if expressed in polar coordinates.