differencial equation

**cinder** · Jun 26th 2007, 09:35 PM

Can someone give me a starting point?

**topsquark** · Jun 26th 2007, 10:08 PM

Originally Posted by cinder

Can someone give me a starting point?

Here's the first part:

$\frac{dV}{dt} = -0.03V$

$\frac{dV}{V} = -0.03dt$

$\int_{V_0}^V \frac{dV}{V} = -0.03 \int_0^t~dt$

$ln(V)|_{V_0}^V = -0.03t|_0^t$

$ln(V) - ln(V_0) = -0.03t$

$ln \left ( \frac{V}{V_0} \right ) = -0.03t$

$V = V_0e^{-0.03t}$

-Dan

**Jhevon** · Jun 26th 2007, 10:48 PM

Originally Posted by cinder

Can someone give me a starting point?

Here's the second part.

Dan provided the equation for V(t)

It is $V(t) = V_0 e^{-0.03t}$

Now we want t such that V(t) is 25% of it's original value, in other words, we want t when

$V(t) = 0.25V_0$

So our equation becomes

$0.25V_0 = V_0 e^{-0.03t}$

$\Rightarrow 0.25 = e^{-0.03t}$

$\Rightarrow t = - \frac {\ln 0.25}{0.03}$

**cinder** · Jun 26th 2007, 11:25 PM

Thanks for the help, I think I got it!

Thread: differencial equation

LinkBack

Thread Tools

Display