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initial value problem

find_the_fun

Active member
Feb 1, 2012
166
\(\displaystyle x=c_1\cos{t}+c_2\sin{t}\) is a two-parameter family of solutions of the DE \(\displaystyle x''+x=0\) Find a solution of the IVP consisting of this differential equation and the following initial conditions:
\(\displaystyle x(\frac{\pi}{6})=\frac{1}{2}\) 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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You can greatly simplify matters be observing that \(\displaystyle \frac{\pi}{6}\) is a special angle. What are:

\(\displaystyle \sin\left(\frac{\pi}{6} \right)\)

\(\displaystyle \cos\left(\frac{\pi}{6} \right)\) ?

You have verified that the given solution is valid by demonstrating that:

\(\displaystyle x''(t)+x(t)=0\)

So, what you want to do (as you did, but use the values of the trig. functions you find above) is generate the 2X2 system of equations from:

\(\displaystyle x\left(\frac{\pi}{6} \right)=\frac{1}{2}\)

\(\displaystyle x'\left(\frac{\pi}{6} \right)=0\)
 

find_the_fun

Active member
Feb 1, 2012
166
You can greatly simplify matters be observing that \(\displaystyle \frac{\pi}{6}\) is a special angle. What are:

\(\displaystyle \sin\left(\frac{\pi}{6} \right)\)

\(\displaystyle \cos\left(\frac{\pi}{6} \right)\) ?

[/MATH]
Ah my calculator wasn't in rad mode.

\(\displaystyle \sin{\frac{\pi}{6}}=\frac{1}{2}\)
\(\displaystyle \cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This is really one of those angles for which you should know the values of the trig. functions. :D

So, using these values, what system of equations do you obtain in the two parameters using the given initial values?
 

find_the_fun

Active member
Feb 1, 2012
166
\(\displaystyle \frac{1}{2}=c_1\frac{\sqrt{3}}{2}+\frac{c_2}{2}\) which can be rewritten as \(\displaystyle c_2=1-c_1\sqrt{3}\)

and from the other equation

\(\displaystyle 0=\frac{-c_1}{2}+\frac{c_2 \sqrt{3}}{2}\) which can be rewritten as \(\displaystyle c_1=c_2\sqrt{3}\)

This gives \(\displaystyle c_2=1-(c_2\sqrt{3})\sqrt{3}\) therefore \(\displaystyle c_2=\frac{1}{4}\) and \(\displaystyle c_1=\frac{\sqrt{3}}{4}\)

By the way I won't be offended if there's something I should really know, so tell me.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You did fine, I got the same values for the parameters. (Sun)