y ′ ′ + 4 y = 0 y''+4y=0 y ′′ + 4 y = 0 The general solution is
y = c 1 sin ( 2 x ) + c 2 cos ( 2 x ) y=c_1\sin (2x)+c_2 \cos(2x) y = c 1 sin ( 2 x ) + c 2 cos ( 2 x ) Given y ( π / 8 ) = 0 y(\pi/8)=0 y ( π /8 ) = 0
0 = c 1 sin ( 2 ( π / 8 ) ) + c 2 cos ( 2 ( π / 8 ) ) 0=c_1\sin (2(\pi/8))+c_2 \cos(2(\pi/8)) 0 = c 1 sin ( 2 ( π /8 )) + c 2 cos ( 2 ( π /8 ))
c 1 + c 2 = 0 c_1+c_2=0 c 1 + c 2 = 0
c 1 = − c 2 c_1=-c_2 c 1 = − c 2 Given y ( π / 6 ) = 1 y(\pi/6)=1 y ( π /6 ) = 1
1 = c 1 sin ( 2 ( π / 6 ) ) + c 2 cos ( 2 ( π / 6 ) ) 1=c_1\sin (2(\pi/6))+c_2 \cos(2(\pi/6)) 1 = c 1 sin ( 2 ( π /6 )) + c 2 cos ( 2 ( π /6 ))
c 1 ( 3 2 ) + c 2 ( 1 2 ) = 1 c_1(\dfrac{\sqrt{3}}{2})+c_2(\dfrac{1}{2})=1 c 1 ( 2 3 ) + c 2 ( 2 1 ) = 1
3 c 1 + c 2 = 2 \sqrt{3}c_1+c_2=2 3 c 1 + c 2 = 2 Then
c 2 = − c 1 c_2=-c_1 c 2 = − c 1
3 c 1 − c 1 = 2 \sqrt{3}c_1-c_1=2 3 c 1 − c 1 = 2
c 1 = 2 3 − 1 c_1=\dfrac{2}{\sqrt{3}-1} c 1 = 3 − 1 2
c 1 = 3 + 1 c_1=\sqrt{3}+1 c 1 = 3 + 1
c 2 = − ( 3 + 1 ) c_2=-(\sqrt{3}+1) c 2 = − ( 3 + 1 ) The solution to the boundary value problem is
y = ( 3 + 1 ) sin ( 2 x ) − ( 3 + 1 ) cos ( 2 x ) y=(\sqrt{3}+1)\sin(2x)-(\sqrt{3}+1)\cos(2x) y = ( 3 + 1 ) sin ( 2 x ) − ( 3 + 1 ) cos ( 2 x )
y = ( 3 + 1 ) ( sin ( 2 x ) − cos ( 2 x ) ) y=(\sqrt{3}+1)\big(\sin(2x)-\cos (2x)\big) y = ( 3 + 1 ) ( sin ( 2 x ) − cos ( 2 x ) )
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