MAT244-2013S > Term Test 1

TT1--Problem 4

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Victor Ivrii:
Find  solution
\begin{equation*}
y^{(4)}+8y''+16y=0
\end{equation*}
satisfying initial conditions
\begin{equation*}
y(0)=1,\; y'(0)=y''(0)=y'''(0)=0.
\end{equation*}

Matthew Cristoferi-Paolucci:
heres my solution

Jason Hamilton:
let y=e^rx

=> (r^4) +8(r^2) =16=0

(r^2 +4)^2 =0

roots= 2i, 2i, -2i, -2i

for double roots y2 and y4: y2=xy1and y4=xy3

y=c1 cos(2x) + c2 sin(2x) +c3 xcos(2x) + c4 xsin(2x)

solve I.C: y(o)=1 => c1=1
y'(0)=y''(0)=y(0)'''=0   =>  c2=c3=0 , c4=1

y=cos(2x) + xsin(2x)

Changyu Li:
$$r^4 + 8 r^2 + 16 = 0 \\ r = \pm 2 i, \pm 2 i\\ y = c_1 e^{2i x} + c_2 e^{-2i x} + c_3 x e^{2i x} + c_4 x e^{-2i x}\\ y' = 2 i c_1 e^{2 i x}-2 i c_2 e^{-2 i x}+c_3 e^{2 i x}+2 i c_3 e^{2 i x} x+c_4 e^{-2 i x}-2 i c_4 e^{-2 i x} x \\ y'' = -4 c_1 e^{2 i x}-4 c_2 e^{-2 i x}+c_3 \left(4 i e^{2 i x}-4 e^{2 i x} x\right)+c_4 \left(-4 i e^{-2 i x}-4 e^{-2 i x} x\right)\\ y''' = -8 i c_1 e^{2 i x}+8 i c_2 e^{-2 i x}+c_3 \left(-8 i e^{2 i x} x-12 e^{2 i x}\right)+c_4 \left(8 i e^{-2 i x} x-12 e^{-2 i x}\right)\\ c_1+c_2=1 \\ 2 i c_1-2 i c_2+c_3+c_4=0 \\ -4 c_1-4 c_2+4 i c_3-4 i c_4=0\\ -8 i c_1+8 i c_2-12 c_3-12 c_4=0\\ c_1 = \frac{1}{2}, c_2 = \frac{1}{2}, c_3 = \frac{-i}{2}, c_4 = \frac{i}{2} \\ y = \frac{1}{2} e^{2 i x}+\frac{1}{2} e^{-2 i x} + \frac{1}{2} i x e^{2 i x} - \frac{1}{2} i x e^{-2 i x}$$

Devin Jeanpierre:

--- Quote from: Jason Hamilton on February 13, 2013, 11:24:39 PM ---for double roots y2 and y4: y2=xy1and y4=xy3

y=c1 cos(2x) + c2 sin(2x) +c3 xcos(2x) + c4 xsin(2x)

solve I.C: y(o)=1 => c1=1
y'(0)=y''(0)=y(0)'''=0   =>  c2=c3=0 , c4=1

y=cos(2x) + xsin(2x)

--- End quote ---
Aw man, using the real solutions / trigonometric decomposition would've really sped things up. I wish I'd thought of that. Clever thinking, dude!