Find a solution for a second-order IVP

Question

The information given:

$$ x = c_{1}cost + c_{2}sint $$

as a two-parameter family of solutions of the second-order DE

$$ x'' + x = 0 $$

Find a solution of the second-order IVP consisting of this DE and the given initial conditions:

$$ x(\frac{\pi}3) = \frac{\sqrt{3}}2 $$ $$ x'(\frac{\pi}3) = 0 $$

My work:

I have plugged in the x appropriately and basically have:

$$ \sqrt{3} = c_{1} + c_{2}\sqrt{3} $$

Which to me, I dont see how to solve it further as there are still two variables and not enough obvious information to go further. What step am I missing or not doing?

Thanks!

Answer 1

You are given the initial conditions $x(\pi/3)=\sqrt 3/2$ and $x'(\pi/3)=0$. Now substitute these into the given solution and its derivative: $$x=c_1\cos t+c_2\sin t$$ $$x'=-c_1\sin t+c_2 \cos t$$ You get a system of two equations involving $c_1$ and $c_2$: $$\begin{cases}\frac{\sqrt 3}{2}&=c_1 \cos (\pi/3)+c_2\sin(\pi/3)\\ 0&=-c_1 \sin(\pi/3)+c_2 \cos(\pi/3)\end{cases}$$ which can be solved.

Answer 2

You have to use the second IC given as well.

Take the derivative, substitute $t=\pi/3$, and it equals zero.

This is the second equations you need to solve with the equation you obtained simultaneously, to find $c_1$ and $c_2$.

$$x'=-c_1\sin t+c_2\cos t$$ $$x'(\pi/3)=-\frac{\sqrt{3}}{2}c_1+\frac{1}{2}c_2=0$$ $$-\sqrt{3}c_1+c_2=0$$