System of Differential Equation with initial conditions

Question

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.

Answer 1

$$ \frac{du}{dt}=\frac{-1}{t_u} u\\ \frac{dw}{dt}=\frac{x}{t_w} - (\frac{1}{t_w}+u)w $$

$$ u = C_1 e^{-t/t_u}\\ \frac{dw}{dt}=\frac{x}{t_w} - (\frac{1}{t_w}+C_1 e^{-t/t_w})w $$

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.