1. ## [SOLVED] Seperable DE

I suppose this is probably pretty easy for people in the know

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...

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

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

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

then I integrated both sides separately

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

leaving me with

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

However wolfram alpha is giving a rather more complicated answer of

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

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

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

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...

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

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

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

then I integrated both sides separately

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

leaving me with

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

However wolfram alpha is giving a rather more complicated answer of

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

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...

3. 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.

4. 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

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

we have that

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

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

Welcome to the world of automated computer algebra!

5. lol, well thanks for clarifying