# First order DE

• Feb 28th 2010, 04:27 PM
arbolis
First order DE
I solved the DE $x'=\frac{1}{t}$. I got $x=\ln (t)+C$.
I've been asked to determine $C$ if $t=-1$.
So I think that the solution of the DE should be $x=\ ln |t|+C$ so that I can reach $C=2$.
I just want to be sure we can talk about negative values of $t$.
• Feb 28th 2010, 04:40 PM
Chris L T521
Quote:

Originally Posted by arbolis
I solved the DE $x'=\frac{1}{t}$. I got $x=\ln (t)+C$.
I've been asked to determine $C$ if $t=-1$.
So I think that the solution of the DE should be $x=\ ln |t|+C$ so that I can reach $C=2$.
I just want to be sure we can talk about negative values of $t$.

Yes, you can do that because $\int\frac{\,dx}{x}=\ln{\color{red}|}x{\color{red}| }+C$. With that, you're able to find the constant C.
• Mar 1st 2010, 02:54 AM
Orbent
Negative?
Your t can be negative, because the anti derivative of 1/t is ln|t| + C. The variable is an absolute value, which of course means that if it is negative you just make it positive :)

I am not entirely sure how to solve this problem because you did not give an x component to go with the t. Usually you are given x, or are can find it out. The ln|-1| is 0, so that would mean C would equal whatever x value you find. Hope that explains it :)
• Mar 1st 2010, 06:56 AM
arbolis
Quote:

Originally Posted by Orbent
Your t can be negative, because the anti derivative of 1/t is ln|t| + C. The variable is an absolute value, which of course means that if it is negative you just make it positive :)

I am not entirely sure how to solve this problem because you did not give an x component to go with the t. Usually you are given x, or are can find it out. The ln|-1| is 0, so that would mean C would equal whatever x value you find. Hope that explains it :)

Yes sorry! I forgot to say that I was given $x(-1)=2$. Hence my value of C. Now I know why the absolute value is extremely important, I won't forget it ever.