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**Matilde****Member**- Registered: 2005-10-30
- Posts: 6

How can I solve this:

y + 9y' + 36y= 0

λ^2 + 9λ + 36= 0 → λ= ?

Matilde

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**Matilde****Member**- Registered: 2005-10-30
- Posts: 6

Hi Mathsyperson!

Thanks for answering me!

I know that the answer is e^((-9/2)x) (Acos(3√7/2)x) + Bsin(3√7/2)x)

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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