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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

\(\displaystyle x(\frac{\pi}{6})=\frac{1}{2}\) and \(\displaystyle x'\frac{\pi}{6}=0\)

So \(\displaystyle x'=c_2\cos{t}-c_1\sin{t}\)

\(\displaystyle x''=-c_2\sin{t}-c_1\cos{t}\)

First thing I'm confused about, I tried to simplify by plugging these into the DE to get.

So \(\displaystyle -c_2\sin{t}-c_1\cos{t}+c_1 \cos{t}+c_2\sin{t}\) which cancels each other off. Why does this happen, what am I doing wrong, how is this not a problem?

Let's try a different approach and solve for \(\displaystyle c_1\) or \(\displaystyle c_2\).

We've got \(\displaystyle x=c_1\cos{t}+c_2\sin{t}\)

\(\displaystyle \frac{1}{2}=c_1\cos{\frac{\pi}{6}}+c_2\sin{\frac{pi}{6}}\)

Solving for \(\displaystyle c_1=\frac{-c_2 \sin{ \frac{\pi}{6}}}{2 \cos{\frac{\pi}{6}}}\)

for the other initial condition we've got \(\displaystyle -c_1\sin{\frac{\pi}{6}}+c_2\cos{\frac{\pi}{6}}=0\)

\(\displaystyle c_2=\frac{c_1 \sin{\frac{\pi}{6}}}{\cos{\frac{\pi}{6}}}\)

Substituting in we've got \(\displaystyle c_1=\frac{\frac{-c_1\sin{\frac{\pi}{6}}}{\cos{frac{\pi}{6}}}}\sin{\frac{\pi}{6}}}}{2\cos\frac{\pi}{6}}}\) which is really ugly.