# Thread: First Order, Linear, PDE

1. ## First Order, Linear, PDE

$\displaystyle u_x + u_y + u = 1$
$\displaystyle u = sin(x)$ on $\displaystyle y = x + x^2, x>0$
Let $\displaystyle x = s$ be the parameter along the initial line $\displaystyle y = x + x^2$

Solution:

Characteristic Equation:
$\displaystyle \frac{dx}{dt} = 1, x = t + s$
$\displaystyle \frac{dy}{dt} = 1, y = t + s +s^2$
$\displaystyle \frac{du}{dt} = 1 - u, u(0) = sin(s)$

Hence:

$\displaystyle y = s + x^2$
$\displaystyle u = 1 + (sin(s) - 1)e^{-t}$

and the solution continues on...
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What I don't understand is how to get $\displaystyle u = 1 + (sin(s) - 1)e^{-t}$.

I sort of understand the $\displaystyle \frac{dx}{dt} = 1, x = t + s$ part, but can someone give me an explanation as well?

Thanks =]

2. $\displaystyle \displaystyle{\frac{du}{1-u}=dt}$

$\displaystyle \displaystyle{\frac{d(1-u)}{1-u}=-dt}$

$\displaystyle ln(1-u)=-t+C$

$\displaystyle 1-u=Ce^{-t}$

$\displaystyle **(1)** \; \: \: u=1-Ce^{-t}$

$\displaystyle u(0)=sin(s)$

$\displaystyle u(0)=sin(s)=1-C$

$\displaystyle C=1-sin(s)$

$\displaystyle Inserting \; to \; (1)$

$\displaystyle u=1-(1-sin(s))e^{-t}$