Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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mathsyperson
2005-11-15 03:58:03

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.

Matilde
2005-11-15 00:07:48

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

mathsyperson
2005-11-12 03:26:13

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)

Matilde
2005-11-12 01:09:26

How can I solve this:

y’’ + 9y' + 36y= 0

λ^2 + 9λ + 36= 0 → λ= ?

Matilde