2nd Order Linear O.D.E. Help

**jwade456** · October 16th 2008, 01:07 AM

Hi guys, I've got this differential equation:

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

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

$Acos{x}+Bsin{x}$

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

Thanks in advance

**Peritus** · October 16th 2008, 02:15 AM

Method of undetermined coefficients - Wikipedia, the free encyclopedia

**Jhevon** · October 16th 2008, 05:09 PM

Originally Posted by jwade456

Hi guys, I've got this differential equation:

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

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

$Acos{x}+Bsin{x}$

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

Thanks in advance

no, your particular solution must be of the form $A \sin \alpha x + B \cos \alpha x$

**mr fantastic** · October 16th 2008, 05:13 PM

Originally Posted by Jhevon

no, your particular solution must be of the form $A \sin \alpha x + B \cos \alpha x$

Unless $\alpha = 1$ ........

**Jhevon** · October 16th 2008, 05:33 PM

Originally Posted by mr fantastic

Unless $\alpha = 1$ ........

it should work even if $\alpha = 1$ , shouldn't it?

**mr fantastic** · October 16th 2008, 06:03 PM

Originally Posted by Jhevon

it should work even if $\alpha = 1$ , shouldn't it?

Have you tried it ......

**Jhevon** · October 16th 2008, 06:54 PM

Originally Posted by mr fantastic

Have you tried it ......

haha, no. but that wink makes me want to take your word for it

**mr fantastic** · October 16th 2008, 09:32 PM

Originally Posted by Jhevon

haha, no. but that wink makes me want to take your word for it

lol! Quite wise. When $\alpha = 1$ the proposed form of particular solution is actually the homogenous solution, that is, the solution to $\ddot{y}+y = 0$ .

**U-God** · October 18th 2008, 08:38 PM

So when $\alpha = 1$ 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.

**mr fantastic** · October 18th 2008, 08:45 PM

Originally Posted by U-God

So when $\alpha = 1$ what form would you have to take to solve the equation?

[snip]

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

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.

Originally Posted by U-God

[snip]
or is there a form that will work regardless of alpha's value?

Cheers.

No.

**U-God** · October 18th 2008, 08:52 PM

Thanks mr fantastic, you certainly live up to your name!!

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

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

Thanks again!

**mr fantastic** · October 18th 2008, 08:57 PM

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: $(x + Cx^2) (A \cos x + B \sin x)$ 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 .....

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

That doesn't work so try the next simplest: $(x + C x^2) (A \cos x + B \sin x)$ . 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.

**U-God** · October 18th 2008, 08:59 PM

You have helped soooo much. Thanks again.

**Chris L T521** · October 18th 2008, 09:05 PM

Originally Posted by mr fantastic

Chris likes D.E.'s ... he might get a kick out of doing it.

Haha

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

I'll take a look at it later.

--Chris

**U-God** · October 19th 2008, 10:23 PM

Originally Posted by mr fantastic

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

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

That doesn't work so try the next simplest: $(x + C x^2) (A \cos x + B \sin x)$ . 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 $x (A \cos x + B \sin x)$ .
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: $x (A \cos x + B \sin x)$ and lo and behold I got the same answer as with form: $(x + C x^2) (A \cos x + B \sin x)$ .

Does that mean I did something wrong?
Cheers,

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