Question #292623

find a solution to the boundary value problem y²+4y=0 y(π/8)=0, y(π/6)=1 if the general solution to the differential equation is y(x)=c1sin2x + c2cos2x

1
Expert's answer
2022-02-01T16:47:43-0500
y+4y=0y''+4y=0

The general solution is


y=c1sin(2x)+c2cos(2x)y=c_1\sin (2x)+c_2 \cos(2x)

Given y(π/8)=0y(\pi/8)=0


0=c1sin(2(π/8))+c2cos(2(π/8))0=c_1\sin (2(\pi/8))+c_2 \cos(2(\pi/8))

c1+c2=0c_1+c_2=0

c1=c2c_1=-c_2

Given y(π/6)=1y(\pi/6)=1


1=c1sin(2(π/6))+c2cos(2(π/6))1=c_1\sin (2(\pi/6))+c_2 \cos(2(\pi/6))

c1(32)+c2(12)=1c_1(\dfrac{\sqrt{3}}{2})+c_2(\dfrac{1}{2})=1

3c1+c2=2\sqrt{3}c_1+c_2=2

Then


c2=c1c_2=-c_1

3c1c1=2\sqrt{3}c_1-c_1=2

c1=231c_1=\dfrac{2}{\sqrt{3}-1}

c1=3+1c_1=\sqrt{3}+1

c2=(3+1)c_2=-(\sqrt{3}+1)

The solution to the boundary value problem is


y=(3+1)sin(2x)(3+1)cos(2x)y=(\sqrt{3}+1)\sin(2x)-(\sqrt{3}+1)\cos(2x)

y=(3+1)(sin(2x)cos(2x))y=(\sqrt{3}+1)\big(\sin(2x)-\cos (2x)\big)


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