First Order DE-Separating the Variables

**bugatti79** · January 18th 2011, 12:32 PM

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

**Ackbeet** · January 18th 2011, 12:34 PM

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

**dwsmith** · January 18th 2011, 12:36 PM

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

**bugatti79** · January 18th 2011, 01:26 PM

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..

**bugatti79** · January 18th 2011, 01:28 PM

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

**dwsmith** · January 18th 2011, 01:34 PM

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.

**dwsmith** · January 18th 2011, 01:38 PM

Originally Posted by dwsmith

Oh you have a negative.

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

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

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

**bugatti79** · January 18th 2011, 01:46 PM

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.

**dwsmith** · January 18th 2011, 01:49 PM

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?

**bugatti79** · January 18th 2011, 02:02 PM

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

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