Higher Order Differential Equation

[Destroxia]

Homework Statement

$y^{(4)} + y = 0, y(0)=0, y'(0)=0,y''(0)=-1,y'''(0)=0$

My issue with this equation is not with the steps, I don't believe but the solving of the IVP, the derivatives of my solution end up being close to 32 terms long, and I was wondering if there is any shorter method I could use without having to figure out the close to 50 terms for all 3 derivatives combined.

Homework Equations

The Attempt at a Solution

(ignore the $c_3$ next to my $c_4$ term, that was a typo)

 

[Dr. Courtney]

http://www.wolframalpha.com/input/?i=y''''+y=0,y(0)=0,y′(0)=0,y′′(0)=−1,y′′′(0)=0

 

[Destroxia]

http://www.wolframalpha.com/input/?i=y''''+y=0,y(0)=0,y′(0)=0,y′′(0)=−1,y′′′(0)=0

Hahah, yes I've tried that already, I'm just pretty sure the professor wants some work shown.

 

[Dr. Courtney]

You can always plug it back in and show with pencil and paper that the solution satisfies the original diff eq and all the initial conditions.

 

[SteamKing]

You can always plug it back in and show with pencul and panper that the solution satisfies the original diff eq and all the initial conditions.

"pencul and panper"?

 

[Dr. Courtney]

Sorry about that. Typing too fast with too little coffee.

 

[HallsofIvy]

"pencul and panper"?

Dr. Courtney is extremely old- "pencul and panper" were used before pencil and paper were invented.

 

[vela]

My issue with this equation is not with the steps, I don't believe but the solving of the IVP, the derivatives of my solution end up being close to 32 terms long, and I was wondering if there is any shorter method I could use without having to figure out the close to 50 terms for all 3 derivatives combined.

Combine terms as you go along and look for patterns. For example, consider the terms $f(x) = c_1 e^{ax}\cos ax + c_2 e^{ax}\sin ax$. First, factor the exponential out so you only have to do the product rule once. When you differentiate, you get
\begin{align*}
f'(x) &= a e^{ax}(c_1 \cos ax + c_2\sin ax) + e^{ax}(-a c_1 \sin ax + a c_2 \cos ax) \\
&= a e^{ax} [(c_1+c_2)\cos ax + (c_2-c_1)\sin ax].
\end{align*} Note that this is essentially the same form you started with, so you can easily write down the second derivative without much effort:
\begin{align*}
f''(x) &= a^2 e^{ax} [((c_1+c_2)+(c_2-c_1)) \cos ax + ((c_2-c_1)-(c_1+c_2))\sin ax] \\
&= a^2 e^{ax} [2c_2 \cos ax - 2c_1 \sin ax] \\
&= 2 a^2 e^{ax} [c_2 \cos ax - c_1 \sin ax]
\end{align*} So in comparison to f(x), differentiating twice produced an overall factor of $2a^2$, changed $\sin x$ into $\cos x$, and changed $\cos x$ into ${-\sin x}$. You should be able to convince yourself that $f''''(x) = -4a^4 e^{ax}(c_1 \cos ax + c_2 \sin ax)$.