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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.

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  • $\begingroup$ You can also use the first condition directly using the known roots of the sine in $y(x)=C\sin(2(x-\frac\pi8))=C\sin(2x-\frac\pi4)$. $\endgroup$ – LutzL Mar 8 '17 at 15:21
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You have a system of 2 -equations to 2 unknowns!! $C1+ C2 = 0$ and thus $C1 = -C2$.

Substitute in the second equations and solve!!

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  • $\begingroup$ That doesn't really explain how you got $C_1+C_2=0$. $\endgroup$ – Kevin Long Mar 8 '17 at 15:21
  • $\begingroup$ @KevinLong : Multiply the first condition with $\sqrt2$. $\endgroup$ – LutzL Mar 8 '17 at 15:22

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