# First Order DE-Separating the Variables

• January 18th 2011, 01:32 PM
bugatti79
First Order DE-Separating the Variables
All,

Going through a book question. Getting stuck right at the end given the IC's See attached... Are the very last two terms equivalent? If so...how?

Any tips appreciated.

Thanks
• January 18th 2011, 01:34 PM
Ackbeet
Could you please type up the DE and the IC? I'm having a very hard time reading that scan.
• January 18th 2011, 01:36 PM
dwsmith
Quote:

Originally Posted by bugatti79
All,

Going through a book question. Getting stuck right at the end given the IC's See attached... Are the very last two terms equivalent? If so...how?

Any tips appreciated.

Thanks

$\displaystyle\int\frac{1}{1+e^{-x}}dx=\ln{(e^x+1)$

You have $x+\ln{(e^x+1)$
• January 18th 2011, 02:26 PM
bugatti79
Quote:

Originally Posted by dwsmith
$\displaystyle\int\frac{1}{1+e^{-x}}dx=\ln{(e^x+1)$

You have $x+\ln{(e^x+1)$

No, its $x+\ln{(e^{-x}+1)$. It works out the same towards the end when you take log of both sides. I keep getting the same answer and including your answer. I wonder is the book wrong? See attached..
• January 18th 2011, 02:28 PM
bugatti79
Quote:

Originally Posted by Ackbeet
Could you please type up the DE and the IC? I'm having a very hard time reading that scan.

Sorry, I have just read you post after sending it. I will do the next time. Sorry
• January 18th 2011, 02:34 PM
dwsmith
Quote:

Originally Posted by bugatti79
No, its $x+\ln{(e^{-x}+1)$. It works out the same towards the end when you take log of both sides. I keep getting the same answer and including your answer. I wonder is the book wrong? See attached..

Oh you have a negative.
• January 18th 2011, 02:38 PM
dwsmith
Quote:

Originally Posted by dwsmith
Oh you have a negative.

$\displaystyle\frac{\cos{y}}{1+e^{-x}}=Ae^x$

$\displaystyle\frac{\cos{\frac{\pi}{4}}}{1+1}=A$

$\displaystyle\frac{\sqrt{2}}{4}=A$
• January 18th 2011, 02:46 PM
bugatti79
Quote:

Originally Posted by dwsmith
$\displaystyle\frac{\cos{y}}{1+e^{-x}}=Ae^x$

$\displaystyle\frac{\cos{\frac{\pi}{4}}}{1+1}=A$

$\displaystyle\frac{\sqrt{2}}{4}=A$

but the book answer is $(1+e^x)Sec y=2\sqrt{2}$
Must be a typo in the book.
• January 18th 2011, 02:49 PM
dwsmith
Quote:

Originally Posted by bugatti79
but the book answer is $(1+e^x)Sec y=2\sqrt{2}$
Must be a typo in the book.

$\displaystyle\cos{y}=\frac{\sqrt{2}}{4}(1+e^x)\Rig htarrow\frac{1}{\cos{y}}=\frac{1}{\frac{\sqrt{2}}{ 4}(1+e^x)}\Rightarrow\sec{y}=\frac{2\sqrt{2}}{1+e^ x}$

See it?
• January 18th 2011, 03:02 PM
bugatti79
Quote:

Originally Posted by dwsmith
$\displaystyle\cos{y}=\frac{\sqrt{2}}{4}(1+e^x)\Rig htarrow\frac{1}{\cos{y}}=\frac{1}{\frac{\sqrt{2}}{ 4}(1+e^x)}\Rightarrow\sec{y}=\frac{2\sqrt{2}}{1+e^ x}$

See it?

I do now, thanks. Just testing the TeXForm function in mathematica to convert the code to latex....

$\sec (y)=\frac{2 \sqrt{2}}{e^x+1}$

THat is so handy! :-)

Cheers