What is the solution for y^{''}+16y=0,y(pi/8)=-2,y^'(pi/8)=16 ?

The solution for y^{''}+16y=0,y(pi/8)=-2,y^'(pi/8)=16 is y=-4cos(4t)-2sin(4t)

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y^{''}+16y=0,y(pi/8)=-2,y^'(pi/8)=16

y^{^{''}} + 16 y = 0, y (\frac{π}{8}) = - 2, y^{^{'}} (\frac{π}{8}) = 16

y = - 4 cos (4 t) - 2 sin (4 t)

Solution steps

y^{''} (t) + 16 y = 0

Solve linear ODE:

y = - 4 cos (4 t) - 2 sin (4 t)

y = - 4 cos (4 t) - 2 sin (4 t)

y^{^{'''}} + 2 y^{^{''}} - 15 y^{^{'}} - 36 y = 0

y^{^{''}} + 8 y^{^{'}} + 6 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

2 y^{^{''}} + 4 y^{^{'}} + 20 y = 0

y^{^{''}} - 4 y^{^{'}} - 96 y = 0, y (0) = 13, y^{^{'}} (0) = 36

y^{^{''}} + 25 y^{^{'}} + \frac{1}{0.046} y = 0

What is the solution for y^{''}+16y=0,y(pi/8)=-2,y^'(pi/8)=16 ?
The solution for y^{''}+16y=0,y(pi/8)=-2,y^'(pi/8)=16 is y=-4cos(4t)-2sin(4t)