Solving the IVP for x=c_1\cos{t}+c_2\sin{t}: Solution to DE & Initial Conditions

[find_the_fun]

[math]x=c_1\cos{t}+c_2\sin{t}[/math] is a two-parameter family of solutions of the DE [math]x''+x=0[/math] Find a solution of the IVP consisting of this differential equation and the following initial conditions:
[math]x(\frac{\pi}{6})=\frac{1}{2}[/math] and [math]x'\frac{\pi}{6}=0[/math]

So [math]x'=c_2\cos{t}-c_1\sin{t}[/math]
[math]x''=-c_2\sin{t}-c_1\cos{t}[/math]

First thing I'm confused about, I tried to simplify by plugging these into the DE to get.
So [math]-c_2\sin{t}-c_1\cos{t}+c_1 \cos{t}+c_2\sin{t}[/math] 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 [math]c_1[/math] or [math]c_2[/math].

We've got [math]x=c_1\cos{t}+c_2\sin{t}[/math]

[math]\frac{1}{2}=c_1\cos{\frac{\pi}{6}}+c_2\sin{\frac{pi}{6}}[/math]
Solving for [math]c_1=\frac{-c_2 \sin{ \frac{\pi}{6}}}{2 \cos{\frac{\pi}{6}}}[/math]

for the other initial condition we've got [math]-c_1\sin{\frac{\pi}{6}}+c_2\cos{\frac{\pi}{6}}=0[/math]
[math]c_2=\frac{c_1 \sin{\frac{\pi}{6}}}{\cos{\frac{\pi}{6}}}[/math]

Substituting in we've got [math]c_1=\frac{\frac{-c_1\sin{\frac{\pi}{6}}}{\cos{frac{\pi}{6}}}}\sin{\frac{\pi}{6}}}}{2\cos\frac{\pi}{6}}}[/math] which is really ugly.

 

[MarkFL]

You can greatly simplify matters be observing that $$\frac{\pi}{6}$$ is a special angle. What are:

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

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

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

$$x''(t)+x(t)=0$$

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:

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

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

 

[find_the_fun]

You can greatly simplify matters be observing that $$\frac{\pi}{6}$$ is a special angle. What are:

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

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

$$

Ah my calculator wasn't in rad mode.

[math]\sin{\frac{\pi}{6}}=\frac{1}{2}[/math]
[math]\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}[/math]

 

[MarkFL]

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]

[math]\frac{1}{2}=c_1\frac{\sqrt{3}}{2}+\frac{c_2}{2}[/math] which can be rewritten as [math]c_2=1-c_1\sqrt{3}[/math]

and from the other equation

[math]0=\frac{-c_1}{2}+\frac{c_2 \sqrt{3}}{2}[/math] which can be rewritten as [math]c_1=c_2\sqrt{3}[/math]

This gives [math]c_2=1-(c_2\sqrt{3})\sqrt{3}[/math] therefore [math]c_2=\frac{1}{4}[/math] and [math]c_1=\frac{\sqrt{3}}{4}[/math]

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

 

[MarkFL]

You did fine, I got the same values for the parameters. (Sun)