Hi,

I'm sure this is straight forward I'm just having a mental block:

I'm trying to solve:

$\displaystyle \frac{dv}{dt}=1+(K+1)v$

should only need the first step or a hint!

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- Apr 13th 2013, 11:12 AMAntSeparable ODE
Hi,

I'm sure this is straight forward I'm just having a mental block:

I'm trying to solve:

$\displaystyle \frac{dv}{dt}=1+(K+1)v$

should only need the first step or a hint! - Apr 13th 2013, 12:10 PMShakarriRe: Separable ODE
There might be a simpler method but you can solve it like in the example here Examples of differential equations - Wikipedia, the free encyclopedia

Where p(x)= -(k+1) and q(x)= 1 - Apr 13th 2013, 12:30 PMAntRe: Separable ODE
Thanks, my lecture notes said it was separable, but I'm pretty sure it's actually not!

- Apr 13th 2013, 01:23 PMBobPRe: Separable ODE
The equation can be solved by the variables separable method,

write it as $\displaystyle \int \frac{dv}{1+(K+1)v}=\int dt.$

It integrates directly to

$\displaystyle \frac{1}{K+1}\ln (1+(K+1)v)=t+C.$ - Apr 13th 2013, 03:23 PMShakarriRe: Separable ODE
Lol can't believe I missed that. I've recently been doing a lot of problems with the more complicated method.