MHB Find specific solution to IVP with two parameters

[find_the_fun]

$$x=c_1 cos(t)+c_2sin(t)$$ is a family of solution to $$x''+x=0$$. Given $$x(\frac{\pi}{6})=\frac{1}{2}$$ and $$x'(\frac{\pi}{6})=0$$ find a solution to the second order IVP consisting of this differential equation and the given intial conditions.

The answer key has [math]x=\frac{\sqrt{3}}{4}cos(t)+\frac{1}{4}sin(t)[/math] which isn't what I'm getting.

So using the first initial condition we know [math]\frac{1}{2}=c_1cos\frac{\pi}{6}+c_2sin\frac{\pi}{6}[/math] gives [math]c_2=1-c_1\sqrt{3}[/math] So [math]x=c_1 cost+(1-c_1 \sqrt{3})sin(t)=c_1 cos(t)+sin(t)-c_1\sqrt{3}sin(t)[/math]
and it's derivative is [math]x'=-c_1sin(t)+cos(t)-c_1 \sqrt{3}cos(t)[/math]
Applying$$ x'(\frac{\pi}{6})=0 $$gives [math]0=-c_1 sin \frac{\pi}{6}+cos\frac{\pi}{6}-c+1\sqrt{3}cos\frac{\pi}{6}[/math] which gives [math]c_1=\frac{\sqrt{3}-1}{3}[/math] and then [math]c_2=1-c_1\sqrt{3}=1-\frac{\sqrt{3}-1}{3} \cdot \sqrt{3}[/math] Where am I going wrong?

 

[MarkFL]

If we are given:

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

then we must state:

$$c_1\cos\left(\frac{\pi}{2}\right)+c_2\sin\left(\frac{\pi}{2}\right)=\frac{1}{2}$$

However, this gives us:

$$c_2=\frac{1}{2}$$

Which is not what the answer you cite reflects. Are you certain the problem is copied correctly?

 

[find_the_fun]

If we are given:

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

then we must state:

$$c_1\cos\left(\frac{\pi}{2}\right)+c_2\sin\left(\frac{\pi}{2}\right)=\frac{1}{2}$$

However, this gives us:

$$c_2=\frac{1}{2}$$

Which is not what the answer you cite reflects. Are you certain the problem is copied correctly?

Sorry that was a typo. It was supposed to be [math]x(\frac{\pi}{6})=\frac{1}{2}[/math]

 

[MarkFL]

Sorry that was a typo. It was supposed to be [math]x(\frac{\pi}{6})=\frac{1}{2}[/math]

Okay, then we find that:

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

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

(1) $$\sqrt{3}c_1+c_2=1$$

$$-c_1\sin\left(\frac{\pi}{6}\right)+c_2\cos\left(\frac{\pi}{6}\right)=0$$

$$-\frac{1}{2}c_1+\frac{\sqrt{3}}{2}c_2=0$$

(2) $$\sqrt{3}c_2-c_1=0$$

Equation (2) implies:

(3) $$c_1=\sqrt{3}c_2$$

and so substituting into (1), we obtain:

$$\sqrt{3}\left(\sqrt{3}c_2\right)+c_2=1$$

$$3c_2+c_2=1$$

$$c_2=\frac{1}{4}$$

And so, (3) gives us:

$$c_1=\sqrt{3}\cdot\frac{1}{4}=\frac{\sqrt{3}}{4}$$

 

[find_the_fun]

Okay, then we find that:

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

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

(1) $$\sqrt{3}c_1+c_2=1$$

$$-c_1\sin\left(\frac{\pi}{6}\right)+c_2\cos\left(\frac{\pi}{6}\right)=0$$

$$-\frac{1}{2}c_1+\frac{\sqrt{3}}{2}c_2=0$$

(2) $$\sqrt{3}c_2-c_1=0$$

Equation (2) implies:

(3) $$c_1=\sqrt{3}c_2$$

and so substituting into (1), we obtain:

$$\sqrt{3}\left(\sqrt{3}c_2\right)+c_2=1$$

$$3c_2+c_2=1$$

$$c_2=\frac{1}{4}$$

And so, (3) gives us:

$$c_1=\sqrt{3}\cdot\frac{1}{4}=\frac{\sqrt{3}}{4}$$

So much simpler than what I was trying to do. I somtimes get confused as to whether I should be using the equation for the ODE or the solution itself.

 

[MarkFL]

The ODE itself will give you the general solution when solved, but you were already provided with that. So, all you need do is take the solution, its derivative, and the initial values to come up with two equations in two unknowns. :D