Toronto Math Forum
MAT2442013S => MAT244 MathTests => Term Test 1 => Topic started by: Victor Ivrii 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*}

heres my solution

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)

$$
r^4 + 8 r^2 + 16 = 0 \\
r = \pm 2 i, \pm 2 i\\
y = c_1 e^{2i x} + c_2 e^{2i x} + c_3 x e^{2i x} + c_4 x e^{2i x}\\
y' = 2 i c_1 e^{2 i x}2 i c_2 e^{2 i x}+c_3 e^{2 i x}+2 i c_3 e^{2 i x} x+c_4 e^{2 i x}2 i c_4 e^{2 i x} x \\
y'' = 4 c_1 e^{2 i x}4 c_2 e^{2 i x}+c_3 \left(4 i e^{2 i x}4 e^{2 i x} x\right)+c_4 \left(4 i e^{2 i x}4 e^{2 i x} x\right)\\
y''' = 8 i c_1 e^{2 i x}+8 i c_2 e^{2 i x}+c_3 \left(8 i e^{2 i x} x12 e^{2 i x}\right)+c_4 \left(8 i e^{2 i x} x12 e^{2 i x}\right)\\
c_1+c_2=1 \\
2 i c_12 i c_2+c_3+c_4=0 \\
4 c_14 c_2+4 i c_34 i c_4=0\\
8 i c_1+8 i c_212 c_312 c_4=0\\
c_1 = \frac{1}{2}, c_2 = \frac{1}{2}, c_3 = \frac{i}{2}, c_4 = \frac{i}{2} \\
y = \frac{1}{2} e^{2 i x}+\frac{1}{2} e^{2 i x} + \frac{1}{2} i x e^{2 i x}  \frac{1}{2} i x e^{2 i x}
$$

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!

$\renewcommand{\Re}{\operatorname{Re}}$
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 $e^{x}$.