Differential Equation involving System of Equations

Question

Find a solution to the boundary value problem \begin{align}y''+ 4y &= 0 \\ y\left(\frac{\pi}{8}\right) &=0\\ y\left(\frac{\pi}{6}\right) &= 1\end{align} if the general solution to the differential equation is $y(x) = C_1 \sin(2x) + C_2 \cos (2x)$.

I was able to compute the following equations: \begin{align}C_1 \left(\frac 12\right)\sqrt2 + C_2 \left(\frac 12\right)\sqrt2 &= 0\\ C_1 \left(\frac 12\right)\sqrt3 + C_2 \left(\frac 12\right) &= 1\end{align}

However I am unable to solve the system of equations. The books says the answer is $\frac{2}{\sqrt3 -1}$for $C_1$ and $-C_2$. I am not sure how to go about manipulating the equations to get on variable.

Answer 1

You have a system of 2 -equations to 2 unknowns!! $C1+ C2 = 0$ and thus $C1 = -C2$.

Substitute in the second equations and solve!!

Answer 2

\begin{align*} \frac{\sqrt 2 C_1}{2} + \frac{\sqrt 2 C_1}{2} &= 0 \\ C_1 + C_2 &= 0 \\ C_2 &= -C_1 \\ \frac{\sqrt{3}C_1}{2} + \frac{C_2}{2} &= 1 \\ \sqrt{3}C_1 + C_2 &= 2 \\ \sqrt{3}C_1 - C_1 &= 2 \\ (\sqrt{3} - 1)C_1 &= 2 \\ C_1 &= \frac{2}{(\sqrt{3} - 1)} \\ C_2 &= -\frac{2}{(\sqrt{3} - 1)} \\ \end{align*}

Which is what the book got.