Thread: differencial equation

Can someone give me a starting point?

Originally Posted by cinder

Can someone give me a starting point?

Here's the first part:

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

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

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

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

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

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

$\displaystyle V = V_0e^{-0.03t}$

-Dan

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 $\displaystyle V(t) = V_0 e^{-0.03t}$

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

$\displaystyle V(t) = 0.25V_0$

So our equation becomes

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

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

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

Thanks for the help, I think I got it!