# Math Help - First Order, Linear, PDE

1. ## First Order, Linear, PDE

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

Solution:

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

Hence:

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

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

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

Thanks =]

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

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

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

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

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

$
u(0)=sin(s)
$

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

$
C=1-sin(s)
$

$
Inserting \; to \; (1)
$

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