# Math Help - Probably a simple problem.

1. ## Probably a simple problem.

It has been awhile since my diffy Q classes. This is my problem:

dx/dt=1/(1+t)

x=1 at t=0

I get:

x=C1*ln(1+t)

When I use the intial value of t=0, x=0, which does not agree with my initial conditions. Can you tell me what I'm doing wrong. Yes, I know this is incredibly simple, but what can I say....i'm simple.

Thanks for any help,
Jason

2. $\frac{dx}{dt}=\frac{1}{1+t}$

$dx=\frac{dt}{1+t}$ let's integrate both sides

$\int dx=\int \frac{dt}{1+t}$

$x=\ln|1+t|+c$ and use the condition t=0 x=1

$1=\ln|1+0|+c$

$1=0+c$ so c=1

$x=\ln(1+t)+1$ we can actually ignore the absolute value condition

3. Thanks!

10 years ago I could have done that....unfortunately it seems I learn one new thing and forget an old thing!