System of Differential Equation with initial conditions I am trying to solve the following system of equations: 
$du/dt = -u / tu$
$dw/dt = (x-w)/tw - u*w$
In the above system, $tu,x,tw$ are all constants. 
I need to find the equations for w and u. 
 A: $$
\frac{du}{dt}=\frac{-1}{t_u} u\\
\frac{dw}{dt}=\frac{x}{t_w} - (\frac{1}{t_w}+u)w
$$
$$
u = C_1 e^{-t/t_u}\\
\frac{dw}{dt}=\frac{x}{t_w} - (\frac{1}{t_w}+C_1 e^{-t/t_w})w
$$
Define auxiliary quantities
$$
A = \frac{x}{t_w}\\
g(t) = (\frac{1}{t_w}+C_1 e^{-t/t_w})
$$
so 
$$
\frac{dw}{dt} + g(t) w = A\\
$$
Multiply both sides by $e^{\int_s^t g(z) dz}$
$$
e^{\int_s^t g(z) dz} \frac{dw}{dt} + e^{\int_s^t g(z) dz} g(t) w = A e^{\int_s^t g(z) dz}\\
\frac{d}{dt} (e^{\int_s^t g(z) dz} w) = A e^{\int_s^t g(z) dz}\\
(e^{\int_s^t g(z) dz} w) = \int^t_{s'} A e^{\int_s^t g(z) dz} + C_2\\
w(t) = \frac{\int^t_{s'} A e^{\int_s^t g(z) dz} + C_2}{e^{\int_s^t g(z) dz}}\\
\int g(t) = \int (\frac{1}{t_w}+C_1 e^{-t/t_w}) = \frac{t}{t_w} - C_1 t_w e^{-t/t_w} +C_3
$$
Changing $s$ just changes $C_3$ and changing $s'$ just changes $C_2$. You will have already fixed $C_1$ from the initial conditions on $u$.
You see the answer only depends on $\frac{C_2}{e^{C_3}}$ so that will be fixed by initial conditions on $w$ and choose $C_2$ and $C_3$ otherwise conveniently.
