a crazy Integral...

• August 11th 2007, 10:13 AM
dudinka
a crazy Integral...
Hi!

1. I have no idea how to solve this integral:

http://img59.imageshack.us/img59/9750/que1hi6.gif

2. I've got two solutions from different sources to this DE, what is the difference and why they chose to write the solutions like that (how do I know when to write them as a product or as a sum)?

the equations is:

y''''+y=0

one solutions is:
http://img59.imageshack.us/img59/6232/tap1tv7.jpg

and the other:
http://img113.imageshack.us/img113/7651/tap2ro4.jpg

Thanks! :)
• August 11th 2007, 10:25 AM
Krizalid
Quote:

Originally Posted by dudinka
1. I have no idea how to solve this integral:

http://img59.imageshack.us/img59/9750/que1hi6.gif

Of course you won't have any idea if you don't know who are $\mu,\,g,\,\lambda$ and $c$

Constants?
• August 11th 2007, 11:48 AM
dudinka
yes, thank you.
• August 11th 2007, 01:57 PM
Soroban
Hello, dudinka!

Quote:

$\int\frac{dx}{(mgx + L)^2 - c^2}$

Let: $mgx + L \:=\:c\sec\theta\quad\Rightarrow\quad mg\,dx \:=\:c\sec\theta\tan\theta\,d\theta\quad\Rightarro w\quad dx \:=\:\frac{c}{mg}\sec\theta\tan\theta\,d\theta$
. . and: . $(mgx + L)^2 - c^2 \:=\:c^2\sec^2\!\theta - c^2 \:=\:c^2(\sec^2\!\theta -1) \:=\:c^2\tan^2\!\theta$

Substitute: . $\int\frac{\frac{c}{mg}\sec\theta\tan\theta\,d\thet a}{c^2\tan^2\theta} \:=\:\frac{1}{cmg}\int\frac{\sec\theta}{\tan\theta }\,d\theta \:=\:\frac{1}{cmg}\int\csc\theta\,d\theta$

. . And we have: . $\frac{1}{cmg}\ln|\csc\theta - \cot\theta| + K$

I'll let you back-substitute . . .

• August 11th 2007, 06:31 PM
topsquark
Quote:

Originally Posted by dudinka
2. I've got two solutions from different sources to this DE, what is the difference and why they chose to write the solutions like that (how do I know when to write them as a product or as a sum)?

the equations is:

y''''+y=0

one solutions is:
http://img59.imageshack.us/img59/6232/tap1tv7.jpg

and the other:
http://img113.imageshack.us/img113/7651/tap2ro4.jpg

I have no idea how the second formula came about. It isn't correct.

To solve $y^{\prime \prime \prime \prime} + y = 0$
you need to solve the characteristic equation:
$m^4 + 1 = 0$

This has as a solution the four 4th roots of -1:
$m = \frac{\sqrt{2}}{2} \pm i \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \pm i \frac{\sqrt{2}}{2}$

So the general solution is:
$y(x) = Ae^{\left ( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right ) x} + Be^{\left ( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right ) x} + Ce^{\left ( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right ) x} + De^{\left ( -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right ) x}$

You may modify this form:
$y(x) = Ae^{\frac{\sqrt{2}}{2}x}e^{i \frac{\sqrt{2}}{2} x} + Be^{\frac{\sqrt{2}}{2}x}e^{-i \frac{\sqrt{2}}{2} x} + Ce^{-\frac{\sqrt{2}}{2}x}e^{i \frac{\sqrt{2}}{2} x} + De^{-\frac{\sqrt{2}}{2}x}e^{-i \frac{\sqrt{2}}{2} x}$

$y(x) = e^{\frac{\sqrt{2}}{2}x} \left ( Ae^{i \frac{\sqrt{2}}{2} x} + Be^{-i \frac{\sqrt{2}}{2} x} \right ) + e^{-\frac{\sqrt{2}}{2}x} \left ( Ce^{i \frac{\sqrt{2}}{2} x} + De^{-i \frac{\sqrt{2}}{2} x} \right )$

Finally:
$y(x) = e^{\frac{\sqrt{2}}{2}x} \left ( A^{\prime}cos \left ( \frac{\sqrt{2}}{2} x \right ) + B^{\prime}sin \left ( \frac{\sqrt{2}}{2} x \right ) \right )$ $+ e^{-\frac{\sqrt{2}}{2}x} \left ( C^{\prime}cos \left ( \frac{\sqrt{2}}{2} x \right ) + D^{\prime}sin \left ( \frac{\sqrt{2}}{2} x \right ) \right )$

These two forms represent the most commonly used forms for this kind of solution.

-Dan
• August 12th 2007, 03:42 AM
dudinka
thanks a lot!