# [SOLVED] Seperable DE

• Jun 3rd 2009, 03:44 AM
Rapid_W
[SOLVED] Seperable DE
I suppose this is probably pretty easy for people in the know :p

Recently I've been using Wolfram Alpha, which is an amazing tool, I mainly like it for showing the steps in calculus, however it doesn't want to give the steps for differential equations :(

Anyway, the question is as follows...

$y'=5e^{4x}\sqrt{1-y^2}$

Now I know it's separable so I've put the square root bit under the y'

$\frac{y'}{\sqrt{1-y^2}}=5e^{4x}$

then I integrated both sides separately

$arcsin(y)=\frac{5}{4}e^{4x}+c$

leaving me with

$y=sin({\frac{5}{4}e^{4x}+c})$

However wolfram alpha is giving a rather more complicated answer of

$y=1+2sinh^2(\frac{1}{8}(4c+5ie^{4x}))$

So is WA wrong, or if it isn't, what steps have been taken to get to that answer?
• Jun 3rd 2009, 03:57 AM
Prove It
Quote:

Originally Posted by Rapid_W
I suppose this is probably pretty easy for people in the know :p

Recently I've been using Wolfram Alpha, which is an amazing tool, I mainly like it for showing the steps in calculus, however it doesn't want to give the steps for differential equations :(

Anyway, the question is as follows...

$y'=5e^{4x}\sqrt{1-y^2}$

Now I know it's separable so I've put the square root bit under the y'

$\frac{y'}{\sqrt{1-y^2}}=5e^{4x}$

then I integrated both sides separately

$arcsin(y)=\frac{5}{4}e^{4x}+c$

leaving me with

$y=sin({\frac{5}{4}e^{4x}+c})$

However wolfram alpha is giving a rather more complicated answer of

$y=1+2sinh^2(\frac{1}{8}(4c+5ie^{4x}))$

So is WA wrong, or if it isn't, what steps have been taken to get to that answer?

I can't see anything wrong with your working out.

I suspect you must have entered the original DE into Wolfram wrongly...
• Jun 3rd 2009, 04:05 AM
Rapid_W
Thanks for clarifying my work, but check for yourself

http://www93.wolframalpha.com/input/...))*(5*exp(4x))

I'm almost certian i've entered it corectly, and i'm confident WA is reading the input correctly.
• Jun 3rd 2009, 07:10 AM
the_doc
Your answer and the one by WA are both correct and are simply two different expressions for the same thing but yours is the simpler and hence better one!

Using the identity

$1 + 2 \sinh^2 z \equiv \cosh 2z$

we have that

$y=1+2 \sinh^2(\frac{1}{8}(4c+5ie^{4x}))$
$= \cosh \left( \frac{5}{4} i e^{4x} + c \right)$
$= \cos \left( \frac{5}{4} e^{4x} + d \right)$ since $\cosh iz \equiv \cos z$ and where $c = id$
$= \sin \left( \frac{\pi}{2} + \frac{5}{4} e^{4x} + d \right)$
$= \sin \left(\frac{5}{4} e^{4x} + C_0 \right)$

where $C_0$ is the constant, $c$ , in your expression.

Welcome to the world of automated computer algebra!
• Jun 3rd 2009, 02:04 PM
Rapid_W
lol, well thanks for clarifying :)