# initial value problem

#### find_the_fun

##### Active member
$$\displaystyle x=c_1\cos{t}+c_2\sin{t}$$ 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

Staff member
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
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

Staff member
This is really one of those angles for which you should know the values of the trig. functions.

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
$$\displaystyle \frac{1}{2}=c_1\frac{\sqrt{3}}{2}+\frac{c_2}{2}$$ 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.