# Solve partial differential equation using characteristic method with non-zero right-side

$$\frac{\partial u }{\partial t} + c \frac{\partial u}{\partial x}=e^{2x}$$ $$-\infty < x < \infty , t>0, c>0$$ initial value $u(x,0) = f(x), -\infty< x < \infty$

I've got the result but it's not satisfy the equation when I substitute it.

characteristic equation

$\frac{dt}{ds} = 1 \implies t = s + t_0,$ let $t(0) = 0 ,$ then $t = s$

$\frac{dx}{ds} = c \implies x = cs + x_0 = ct + x_0$

$\frac{du}{ds} = e^{2x} \implies u = s e^{2x} + u_0 = t e^{2x} + u_0$

$u_0 = u \left(x(0),0 \right) = f(x_0) = f(x-ct)$

so, $u = t e^{2x} + f(x-ct)$

is it correct?

-
is it really works? what do you mean by not a fully general solution? – chihiroasleaf Feb 20 '13 at 14:54
If I substitute the result, I get $u_x = 2te^{2x} + f'(x-ct)$ and $u_t = e^{2x} - c f'(x-ct)$ so $u_t + cu_x \neq e^{2x}$ – chihiroasleaf Feb 20 '13 at 15:04
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Apr 21 '13 at 1:26

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=c$ , letting $x(0)=x_0$ , we have $x=cs+x_0=ct+x_0$
$\dfrac{du}{ds}=e^{2x}=e^{2cs+2x_0}$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=\dfrac{e^{2cs+2x_0}-e^{2x_0}}{2c}+f(x_0)=\dfrac{e^{2x}-e^{2x-2ct}}{2c}+f(x-ct)$