# 2nd Order Linear O.D.E. Help

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• Oct 16th 2008, 01:07 AM
2nd Order Linear O.D.E. Help
Hi guys, I've got this differential equation:

$\displaystyle \ddot{y}+y=4sin{\alpha{x}}$

And I've worked out the general solution when it is set equal to zero to be:

$\displaystyle Acos{x}+Bsin{x}$

I need help finding the equation to set y(x) to so that i can solve the rest of it. Any clues?

• Oct 16th 2008, 02:15 AM
Peritus
• Oct 16th 2008, 05:09 PM
Jhevon
Quote:

Originally Posted by jwade456
Hi guys, I've got this differential equation:

$\displaystyle \ddot{y}+y=4sin{\alpha{x}}$

And I've worked out the general solution when it is set equal to zero to be:

$\displaystyle Acos{x}+Bsin{x}$

I need help finding the equation to set y(x) to so that i can solve the rest of it. Any clues?

no, your particular solution must be of the form $\displaystyle A \sin \alpha x + B \cos \alpha x$
• Oct 16th 2008, 05:13 PM
mr fantastic
Quote:

Originally Posted by Jhevon
no, your particular solution must be of the form $\displaystyle A \sin \alpha x + B \cos \alpha x$

Unless $\displaystyle \alpha = 1$ ........
• Oct 16th 2008, 05:33 PM
Jhevon
Quote:

Originally Posted by mr fantastic
Unless $\displaystyle \alpha = 1$ ........

it should work even if $\displaystyle \alpha = 1$, shouldn't it?
• Oct 16th 2008, 06:03 PM
mr fantastic
Quote:

Originally Posted by Jhevon
it should work even if $\displaystyle \alpha = 1$, shouldn't it?

Have you tried it ...... (Wink)
• Oct 16th 2008, 06:54 PM
Jhevon
Quote:

Originally Posted by mr fantastic
Have you tried it ...... (Wink)

haha, no. but that wink makes me want to take your word for it :D
• Oct 16th 2008, 09:32 PM
mr fantastic
Quote:

Originally Posted by Jhevon
haha, no. but that wink makes me want to take your word for it :D

lol! Quite wise. When $\displaystyle \alpha = 1$ the proposed form of particular solution is actually the homogenous solution, that is, the solution to $\displaystyle \ddot{y}+y = 0$.
• Oct 18th 2008, 08:38 PM
U-God
So when $\displaystyle \alpha = 1$ what form would you have to take to solve the equation?

Furthermore, do you have to solve the equation with two different forms depending on whether or not alpha equals one, or is there a form that will work regardless of alpha's value?

Cheers.
• Oct 18th 2008, 08:45 PM
mr fantastic
Quote:

Originally Posted by U-God
So when $\displaystyle \alpha = 1$ what form would you have to take to solve the equation?

[snip]

When $\displaystyle \alpha = 1$, use the form $\displaystyle (x + Cx^2) (A \cos x + B \sin x)$ to construct a particular solution.

Quote:

Originally Posted by U-God
[snip]
Furthermore, do you have to solve the equation with two different forms depending on whether or not alpha equals one,
[snip]

Yes.

Quote:

Originally Posted by U-God
[snip]
or is there a form that will work regardless of alpha's value?

Cheers.

No.
• Oct 18th 2008, 08:52 PM
U-God
Thanks mr fantastic, you certainly live up to your name!!

If I may ask just one more question, where did you get the form: $\displaystyle (x + Cx^2) (A \cos x + B \sin x)$ from?

Is there a way I can derive it myself? Or is it just a rule of thumb.

Thanks again!
• Oct 18th 2008, 08:57 PM
mr fantastic
Quote:

Originally Posted by U-God
Thanks mr fantastic, you certainly live up to your name!!

If I may ask just one more question, where did you get the form: $\displaystyle (x + Cx^2) (A \cos x + B \sin x)$ from?

Is there a way I can derive it myself? Or is it just a rule of thumb.

Thanks again!

The informal answer is that it's the simplest form that works .....

$\displaystyle (A \cos x + B \sin x)$ obviously is no good so try the next simplest: $\displaystyle x (A \cos x + B \sin x)$.

That doesn't work so try the next simplest: $\displaystyle (x + C x^2) (A \cos x + B \sin x)$. That works so it's all you need. Consider it a rule of thumb.

There is a way of getting it formally but I don't have the time to type out the details right now. Chris likes D.E.'s ... he might get a kick out of doing it.
• Oct 18th 2008, 08:59 PM
U-God
You have helped soooo much. Thanks again.
• Oct 18th 2008, 09:05 PM
Chris L T521
Quote:

Originally Posted by mr fantastic
Chris likes D.E.'s ... he might get a kick out of doing it.

Haha (Rofl)

But like you, I don't have the time right now... (Crying)

I'll take a look at it later. (Nod)

--Chris
• Oct 19th 2008, 10:23 PM
U-God
Quote:

Originally Posted by mr fantastic
The informal answer is that it's the simplest form that works .....

$\displaystyle (A \cos x + B \sin x)$ obviously is no good so try the next simplest: $\displaystyle x (A \cos x + B \sin x)$.

That doesn't work so try the next simplest: $\displaystyle (x + C x^2) (A \cos x + B \sin x)$. That works so it's all you need. Consider it a rule of thumb.

There is a way of getting it formally but I don't have the time to type out the details right now. Chris likes D.E.'s ... he might get a kick out of doing it.

Thanks again,

I handed my work in and it all looks good, however some of my friends said that they did it by using the form $\displaystyle x (A \cos x + B \sin x)$.
You said here that that form wouldn't work and to move on to the next form up, so out of interest when I got home I tried it with form: $\displaystyle x (A \cos x + B \sin x)$ and lo and behold I got the same answer as with form: $\displaystyle (x + C x^2) (A \cos x + B \sin x)$.

Does that mean I did something wrong?
Cheers,
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