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#1 2005-11-12 01:09:26

Matilde
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differential equations

How can I solve this:

y + 9y' + 36y= 0

λ^2 + 9λ + 36= 0 → λ= ?  hmm

Matilde

#2 2005-11-12 03:26:13

mathsyperson
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Re: differential equations

It looks like it's going to have complex roots, but we'll see.

x = (-b ± √ (b² - 4ac))÷ 2a

x = (-9 ± √ (-63))÷ 2

The negative square root confirms my suspicions.

The roots work out to be -4.5 ± (1.5√7)i, where i is √(-1).

If the solutions to the auxiliary equation are complex, they go into the following equation:

λ = c±di ∴ y = e^c (Acosdx + Bsindx), where A and B are arbitrary constants.

For your example, y = e^-4.5 (Acos(1.5√7)x + Bsin(1.5√7)x)


Why did the vector cross the road?
It wanted to be normal.

#3 2005-11-15 00:07:48

Matilde
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Re: differential equations

Hi Mathsyperson!
Thanks for answering me! smile
I know that the answer is e^((-9/2)x)  (Acos(3√7/2)x) + Bsin(3√7/2)x) hmm

#4 2005-11-15 03:58:03

mathsyperson
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Re: differential equations

Ooh. The e^4.5 bit was just a carried forward mistake from me copying down the answer to the auxiliary equation wrong. That was stupid. Whenever you saw :pm I meant ± and I just got the code wrong. I've edited my post, so hopefully it makes a bit more sense now.


Why did the vector cross the road?
It wanted to be normal.

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