Higher Order Differential Equation

1. Jul 19, 2015

RyanTAsher

1. The problem statement, all variables and given/known data

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

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.

2. Relevant equations

3. The attempt at a solution

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

2. Jul 19, 2015

Dr. Courtney

3. Jul 19, 2015

RyanTAsher

4. Jul 20, 2015

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.

Last edited: Jul 20, 2015
5. Jul 20, 2015

SteamKing

Staff Emeritus
"pencul and panper"?

6. Jul 20, 2015

Dr. Courtney

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

7. Jul 20, 2015

HallsofIvy

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

8. Jul 20, 2015

vela

Staff Emeritus
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 $f''''(x) = -4a^4 e^{ax}(c_1 \cos ax + c_2 \sin ax)$.

Last edited: Jul 20, 2015