1. ## Integration Help

I have the following direct intergration problem:

dy/dt = v0/(1+kv0t)

y = 1/k In (1+kv0t) + c

I see where the log comes from, just not sure about the 1/k?

2. Originally Posted by woody
I have the following direct intergration problem:

dy/dt = v0/(1+kv0t)

y = 1/k In (1+kv0t) + c

I see where the log comes from, just not sure about the 1/k?

$\frac{dy}{dt} = \frac{v_0}{1 + kv_0t}$

$y = \int{\frac{v_0}{1 + kv_0t}\,dt}$

$y = \frac{1}{k}\int{\frac{kv_0}{1 + kv_0t}\,dt}$.

Let $u= 1 + kv_0t$ so that $\frac{du}{dt} = kv_0$

So the integral becomes

$y = \frac{1}{k}\int{\frac{1}{u}\,\frac{du}{dt}\,dt}$

$= \frac{1}{k}\int{\frac{1}{u}\,du}$

$= \frac{1}{k}\ln{|u|} + C$

$= \frac{1}{k}\ln{|1 + kv_0t|} + C$.

3. Originally Posted by woody
I have the following direct intergration problem:

dy/dt = v0/(1+kv0t)

When you substitute, don't forget to change "dt" also. Let u= 1+ kv0t. The [tex]du= kv_0dt so $dt= \frac{1}{kv_0} du$. Putting that into the integral, $\int \frac{v_0}{1+kv_0t}dt$ becomes $\int \frac{v_0}{u} \frac{1}{kv_0}du= \frac{1}{k}\int \frac{1}{u} du$. The $v_0$ in the numerator cancels the $v_0$ in the denominator but leaves k in the denominator.