How would I use separation of variable to find the general solution to this differential equation? I don't really even know where to start.
$\displaystyle W'(t) = \mu_0 e^{-Dt}W$
Generally specking if we wtite the equation as...
$\displaystyle w^{'} = a(t)\cdot w$ (1)
... we can separate the variables sterting from (1) in such way...
$\displaystyle \frac {dw}{dt} = a(t)\cdot w \rightarrow \frac{dw}{w} = a(t)\cdot dt \rightarrow \ln w = \int a(t)\cdot dt + \ln c$ (2)
... where $\displaystyle c>0$ is an 'arbitrary constant'...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Now I'm really confused.
Where did a(t) come from? Or are you just speaking generally? Should I be assigning aspects of the equation to those variables? e.g:
$\displaystyle a(t) = \mu_0 e^{-Dt}$
Any extra help would be greatly appreciated. I really can't seem to grasp the separation of variable method. Do you know of any good examples that step through working?
Ah, I think I just had an epiphany thanks to Midnight Tutor (MidnightTutor.com Free Online Calculus and Chemistry Tutorials).
$\displaystyle W'(t) = \mu_0 e^{-Dt}W$
$\displaystyle \frac{dW}{dt} = \mu_0 e^{-Dt}W$
$\displaystyle \int{\frac{dW}{dt}} = \int{\mu_0 e^{-Dt}W}$
$\displaystyle lnW = \frac{-\mu_0e^{-Dt}}{D}$
Is this the general solution? or does it need to be taken somewhere from there?
Setting $\displaystyle a(t)= \mu_{0} \cdot e^{-Dt}$ and applying the procedure I have indicated you arrive to write...
$\displaystyle \ln w = - \frac{\mu_{0}}{D} \cdot e^{-Dt} + \ln c$ (1)
... where c>0 is an arbitrary constant. Exponentiating both terms of (1) you obtain...
$\displaystyle w= c\cdot e^{-\frac{\mu_{0}}{D} e^{-Dt}} $ (2)
That's the general solution!...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$