# Thread: Problem with first order separabel differential equation

1. ## Problem with first order separabel differential equation

Hi!

First of all I am Norwegian, and my math-english might not be totally correct.

Okay, I have a problem with a first order separabel differential equation:

where A is a real constant and 0 < y < A.

I seperate:

Splitting the fraction and differentiate it:

I now have to put the y on the one side of the equation, but I'm unsure of how to do this. This is as far as i have gotten:

Is there something I am doing wrong in the steps above?

Any help would be greatly appreciated!

Thomas

2. Originally Posted by luckytommy
Hi!

First of all I am Norwegian, and my math-english might not be totally correct.

Okay, I have a problem with a first order separabel differential equation:

I seperate:

Splitting the fraction and differentiate it:

Mr F says: This should be $\displaystyle {\color{red} \frac{1}{A \lambda} \ln \left(\frac{y}{A {\color{blue}-} y}\right) = t + C}$

I now have to put the y on the one side of the equation, but I'm unsure of how to do this. This is as far as i have gotten:

Is there something I am doing wrong in the steps above?

Any help would be greatly appreciated!

Thomas
$\displaystyle \frac{1}{A \lambda} \ln \frac{y}{A - y} = t + C$

$\displaystyle \Rightarrow \ln \frac{y}{A - y} = A \lambda (t + C) = A \lambda t + K$

$\displaystyle \Rightarrow \frac{y}{A - y} = e^{A \lambda t + K} = D e^{A \lambda t}$

and now your job is to make y the subject.

3. Thank you mr F.

I did some more manipulation, and came to this answer:

$\displaystyle y=Ae^{A\lambda t+C}-ye^{A\lambda t+C}$

$\displaystyle y(1+e^{A\lambda t+C})=Ae^{A\lambda t+C}$

$\displaystyle y=\frac{Ae^{A\lambda t+C}}{1+e^{A\lambda t+C}}$

Correctamente?

I see from your post that you created a new constant D. That is a better way to do it... You really are fantastic aren't you?

Much obliged

4. Originally Posted by luckytommy
Thank you mr F.

I did some more manipulation, and came to this answer:

$\displaystyle y=Ae^{A\lambda t+C}-ye^{A\lambda t+C}$

$\displaystyle y(1+e^{A\lambda t+C})=Ae^{A\lambda t+C}$

$\displaystyle y=\frac{Ae^{A\lambda t+C}}{1+e^{A\lambda t+C}}$

Correctamente?

I see from your post that you created a new constant D. That is a better way to do it... You really are fantastic aren't you?

Much obliged
Correctamente!