# Problem with first order separabel differential equation

• Dec 9th 2009, 03:46 AM
luckytommy
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:

http://folk.uio.no/thomaali/1.jpg where A is a real constant and 0 < y < A.

I seperate:

http://folk.uio.no/thomaali/2.jpg

Splitting the fraction and differentiate it:

http://folk.uio.no/thomaali/3.jpg

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:

http://folk.uio.no/thomaali/4.jpg

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

Any help would be greatly appreciated!

Thomas
• Dec 9th 2009, 04:00 AM
mr fantastic
Quote:

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:

http://folk.uio.no/thomaali/1.jpg

I seperate:

http://folk.uio.no/thomaali/2.jpg

Splitting the fraction and differentiate it:

http://folk.uio.no/thomaali/3.jpg 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:

http://folk.uio.no/thomaali/4.jpg

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.
• Dec 9th 2009, 06:16 AM
luckytommy
Thank you mr F.(Bow)

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 :)
• Dec 9th 2009, 04:29 PM
mr fantastic
Quote:

Originally Posted by luckytommy
Thank you mr F.(Bow)

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!