# Solving ODE's from a PDE

I am currently working on a question and require some help to proceed. The question is:

Use Method of Characteristics for: $$y\cdot u_x-2xy\cdot u_y=2xu$$ With $u=y^3$ when $x=0$ and $1\leq y\leq 2$

Now I am trying to solve it, I have the characteristic equations: $$\dot x(s)=y(s),\>\> \dot y(s) = 2x(s)y(s), \>\> \dot z(s)=2x(s)z(s)$$ I'm not sure if these are the right general auxiliary conditions, but I put: $$x(0)=0, \>\>y(0)=y_0, \>\> z(0)=y_0^3$$ Now to solve for $x(s)$ and $y(s)$, the question says to multiply the first by $2x(s)$ and then add. I'm not sure how that would help, and am looking for some guidance. Thanks!

• Oh yes, $z(s)$ denotes the values of the solution along a characteristic. Forgot people have different representations of such a notation. @JJacquelin – Felicio Grande Sep 18 '17 at 15:17

$$y u_x-2xy u_y=2xu$$

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

A second family of characteristic curves comes from $\quad\frac{dy}{-2xy}=\frac{du}{2xu}\quad\to\quad yu=c_2$

The general solution of the PDE on the form of implicit equation is : $$\Phi\big((y+x^2)\:,\:(yu)\big)=0$$ where $\Phi$ is any differentiable function of two variables.

Equivalently, $u$ can be expressed on explicit form : $$yu=F(y+x^2) \quad\to\quad u(x,y)=\frac{1}{y}F(y+x^2)$$ where $F(X)$ is any differentiable function.

$F(X)$ has to be determined according to the boundary condition.

$u(0,y)=y^3=\frac{1}{y}F(y+0^2) \quad\to\quad y^4=F(y)\quad$ which determines the function $$F(X)=X^4$$ Putting this function into the general solution gives the particular solution: $$u(x,y)=\frac{1}{y}(y+x^2)^4$$