Originally Posted by

**woody198403** I'm doing revision for an exam in 2 days and I don't quite understand one of the final steps in the problem (worked example)

$\displaystyle x u_x + yu_y = ku$ subject to $\displaystyle u(x,1) = F(x)$

Solution

Parametrization of IC's ->

$\displaystyle x(s) = s, y(s) = 1, u(s) = F(s)$

I don't quite understand why this is done. I can see how its used to solve this question....is it just a standard method to solve PDE's? **This is standard to get an ODE.**

$\displaystyle \frac{dx}{dt} = x, \ \frac{dy}{dt} = y$

Then $\displaystyle x = c_1(s)e^t, \ y = c_2(s) e^t$

Apply the initial data to get $\displaystyle x = se^t, \ y = e^t$

$\displaystyle t = \ln \ y$ I'm not sure why this is....(Worried) **You took the logarithm of both sides (you should be worried in a PDE class if this is not clear).**

and $\displaystyle s = \frac{x}{e^t} = \frac{x}{y}$

PDE becomes $\displaystyle \frac{du}{dt} = ku$

$\displaystyle \frac{du}{u} = k \ dt$

and I don't understand any of the remaining steps:

$\displaystyle \ln u = kt + c$ (I'm not sure where $\displaystyle \ln u$ comes from and the RHS is just integrating with respect to t, right? **You took the integral of both sides (first method you learn to solve a DE in Calculus 1). The left side is clearly with respect to u and the right side is with respect to t.**

$\displaystyle u = D(s) e^{kt}$ Im not sure what $\displaystyle D(s)$ is or where the $\displaystyle e^{kt}$ term comes from **You exponentiated both sides and the D(s) comes from the fact that your constant C is a function of s (which might just be a real number).**

$\displaystyle u(s,t = 0) = D(s) = F(s)$. I don't understand this because I don't know what $\displaystyle D(s)$ is. **That's why you plugged in t=0, it gives you that D(s)=F(s), where F(s) is given in the problem.**

$\displaystyle u = F(s) e^{kt} = F \Big( \frac{x}{y}\Big) (e^t)^k = F(\frac{x}{y})y^k$ I actually think I understand this line.

Can someone please explain this to me? I'm worried.