# Thread: First Order DE-Separating the Variables

1. ## 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

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

3. 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
$\displaystyle \displaystyle\int\frac{1}{1+e^{-x}}dx=\ln{(e^x+1)$

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

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

You have $\displaystyle x+\ln{(e^x+1)$
No, its $\displaystyle x+\ln{(e^{-x}+1)$. 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..

5. 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

6. Originally Posted by bugatti79
No, its $\displaystyle x+\ln{(e^{-x}+1)$. 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.

7. Originally Posted by dwsmith
Oh you have a negative.
$\displaystyle \displaystyle\frac{\cos{y}}{1+e^{-x}}=Ae^x$

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

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

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

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

$\displaystyle \displaystyle\frac{\sqrt{2}}{4}=A$
but the book answer is $\displaystyle (1+e^x)Sec y=2\sqrt{2}$
Must be a typo in the book.

9. Originally Posted by bugatti79
but the book answer is $\displaystyle (1+e^x)Sec y=2\sqrt{2}$
Must be a typo in the book.
$\displaystyle \displaystyle\cos{y}=\frac{\sqrt{2}}{4}(1+e^x)\Rig htarrow\frac{1}{\cos{y}}=\frac{1}{\frac{\sqrt{2}}{ 4}(1+e^x)}\Rightarrow\sec{y}=\frac{2\sqrt{2}}{1+e^ x}$

See it?

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

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

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

THat is so handy! :-)

Cheers