Author Topic: TT1--Problem 4  (Read 5018 times)

Victor Ivrii

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TT1--Problem 4
« on: February 13, 2013, 10:41:28 PM »
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

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Re: TT1--Problem 4
« Reply #1 on: February 13, 2013, 11:17:49 PM »
heres my solution
« Last Edit: February 13, 2013, 11:24:22 PM by Matthew Cristoferi-Paolucci »

Jason Hamilton

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Re: TT1--Problem 4
« Reply #2 on: February 13, 2013, 11:24:39 PM »
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

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Re: TT1--Problem 4
« Reply #3 on: February 14, 2013, 12:04:40 AM »
$$
r^4 + 8 r^2 + 16 = 0 \\
r = \pm 2 i, \pm 2 i\\
y = c_1 e^{2i x} + c_2 e^{-2i x} + c_3 x e^{2i x} + c_4 x e^{-2i x}\\
y' = 2 i c_1 e^{2 i x}-2 i c_2 e^{-2 i x}+c_3 e^{2 i x}+2 i c_3 e^{2 i x} x+c_4 e^{-2 i x}-2 i c_4 e^{-2 i x} x \\
y'' = -4 c_1 e^{2 i x}-4 c_2 e^{-2 i x}+c_3 \left(4 i e^{2 i x}-4 e^{2 i x} x\right)+c_4 \left(-4 i e^{-2 i x}-4 e^{-2 i x} x\right)\\
y''' = -8 i c_1 e^{2 i x}+8 i c_2 e^{-2 i x}+c_3 \left(-8 i e^{2 i x} x-12 e^{2 i x}\right)+c_4 \left(8 i e^{-2 i x} x-12 e^{-2 i x}\right)\\
c_1+c_2=1 \\
2 i c_1-2 i c_2+c_3+c_4=0 \\
-4 c_1-4 c_2+4 i c_3-4 i c_4=0\\
-8 i c_1+8 i c_2-12 c_3-12 c_4=0\\
c_1 = \frac{1}{2}, c_2 = \frac{1}{2}, c_3 = \frac{-i}{2}, c_4 = \frac{i}{2} \\
y = \frac{1}{2} e^{2 i x}+\frac{1}{2} e^{-2 i x} + \frac{1}{2} i x e^{2 i x} - \frac{1}{2} i x e^{-2 i x}
$$

Devin Jeanpierre

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Re: TT1--Problem 4
« Reply #4 on: February 14, 2013, 12:06:58 AM »
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)
Aw man, using the real solutions / trigonometric decomposition would've really sped things up. I wish I'd thought of that. Clever thinking, dude!

Victor Ivrii

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Re: TT1--Problem 4
« Reply #5 on: February 14, 2013, 05:40:25 AM »
$\renewcommand{\Re}{\operatorname{Re}}$

Changyu -- wrong signs in the last 2 terms. Then your solution would collapse to $\cos(2x)+x\sin(2x)$ rather than to $\cos(2x)-x\sin(2x)$ as your solution. Definitely if you going through complex form you need to work only with half of roots so you look for
\begin{equation*}
\Re \bigl(C_1 e^{2ix} + C_2 x e^{2ix}\bigr).
\end{equation*}

Because of initial data, the real form is preferable but there is more. Note $y'(0)=y'''(0)=0$ and equation contains only even derivatives. Therefore if $y(x)$ is a solution, $y(-x)$ is also a solution and since solution is unique we conclude that $y(x)=y(-x)$ so it is an even function i.e.
\begin{equation*}
y(x) = C_1 \cos(2x) +C_3 x\sin(2x)
\end{equation*}
and now everything goes very fast: $y(0)=C_1$, $y''(0)=-4C_1+4 C_3$.

Devin, each approach has its own advantages and disadvantages and you need to be comfortable with both. Also note that $\cos(x)$ and $\sin(x)$ are not only real but also even and odd respectively which often makes life easier; $\cosh(x)$ and $\sinh(x)$ are used by the same reason and sometimes they give you an edge over pair $e^x$ and $e^{-x}$.
« Last Edit: February 14, 2013, 06:07:30 AM by Victor Ivrii »