Separation of variables to solve growth equation

**drew.walker** · September 15th 2009, 10:27 PM

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.

$W'(t) = \mu_0 e^{-Dt}W$

**chisigma** · September 15th 2009, 10:56 PM

Generally specking if we wtite the equation as...

$w^{'} = a(t)\cdot w$ (1)

... we can separate the variables sterting from (1) in such way...

$\frac {dw}{dt} = a(t)\cdot w \rightarrow \frac{dw}{w} = a(t)\cdot dt \rightarrow \ln w = \int a(t)\cdot dt + \ln c$ (2)

... where $c>0$ is an 'arbitrary constant'...

Kind regards

$\chi$ $\sigma$

**drew.walker** · September 15th 2009, 11:06 PM

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:

$a(t) = \mu_0 e^{-Dt}$

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?

**drew.walker** · September 15th 2009, 11:22 PM

Ah, I think I just had an epiphany thanks to Midnight Tutor (MidnightTutor.com Free Online Calculus and Chemistry Tutorials).

$W'(t) = \mu_0 e^{-Dt}W$

$\frac{dW}{dt} = \mu_0 e^{-Dt}W$

$\int{\frac{dW}{dt}} = \int{\mu_0 e^{-Dt}W}$

$lnW = \frac{-\mu_0e^{-Dt}}{D}$

Is this the general solution? or does it need to be taken somewhere from there?

**chisigma** · September 15th 2009, 11:42 PM

Setting $a(t)= \mu_{0} \cdot e^{-Dt}$ and applying the procedure I have indicated you arrive to write...

$\ln w = - \frac{\mu_{0}}{D} \cdot e^{-Dt} + \ln c$ (1)

... where c>0 is an arbitrary constant. Exponentiating both terms of (1) you obtain...

$w= c\cdot e^{-\frac{\mu_{0}}{D} e^{-Dt}}$ (2)

That's the general solution!...

Kind regards

$\chi$ $\sigma$

**drew.walker** · September 15th 2009, 11:47 PM

Ah, okay. Now I'm starting to get it. Thanks for the help.

Thread: Separation of variables to solve growth equation

LinkBack

Thread Tools

Display

Separation of variables to solve growth equation

Similar Math Help Forum Discussions

Solving Poissons equation with a delta function in RHS using separation of variables.

homogeneous differential equation->separation of variables

solve the differential equation by separation of variables dy/dx = (cos x)e^(y+sinx)

solving a first order differential equation using separation of variables

need help with separation or variable to solve wave equation

Search Tags