What is the solution for 2y^{''}+16y=0,y(0)=0.125,y^'(0)=0 ?

The solution for 2y^{''}+16y=0,y(0)=0.125,y^'(0)=0 is y=0.125cos(2^{1/2+1}t)

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2y^{''}+16y=0,y(0)=0.125,y^'(0)=0

2 y^{^{''}} + 16 y = 0, y (0) = 0.125, y^{^{'}} (0) = 0

y = 0.125 cos (2^{\frac{1}{2} + 1} t)

Solution steps

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

Solve linear ODE:

y = 0.125 cos (2^{\frac{1}{2} + 1} t)

y = 0.125 cos (2^{\frac{1}{2} + 1} t)

y^{^{''}} + 23 y^{^{'}} + 3 y = 0

y^{^{'''}} - 6 y^{^{''}} - 27 y^{^{'}} = 0

y^{^{''}} + 3 y^{^{'}} - 4 y = 0, y (0) = 7, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 y = 0, y (0) = 2

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 10

What is the solution for 2y^{''}+16y=0,y(0)=0.125,y^'(0)=0 ?
The solution for 2y^{''}+16y=0,y(0)=0.125,y^'(0)=0 is y=0.125cos(2^{1/2+1}t)