What is the solution for 2y^{''}+8y=0,y^'(0)=1,y(0)=0 ?

The solution for 2y^{''}+8y=0,y^'(0)=1,y(0)=0 is y= 1/2 sin(2t)

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2y^{''}+8y=0,y^'(0)=1,y(0)=0

2 y^{^{''}} + 8 y = 0, y^{^{'}} (0) = 1, y (0) = 0

y = \frac{1}{2} sin (2 t)

Solution steps

2 y^{''} (t) + 8 y = 0

Solve linear ODE:

y = \frac{1}{2} sin (2 t)

y = \frac{1}{2} sin (2 t)

(D^{4} - 2 D^{3} - 3 D^{2} + 4 D + 4) y = 0

40 y^{^{''}} + 7 y^{^{'}} - 3 y = 0

8 y^{^{''}} + 8 y^{^{'}} + \frac{1}{4} y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + k^{2} \cdot y = 0

y^{^{''}} = y

What is the solution for 2y^{''}+8y=0,y^'(0)=1,y(0)=0 ?
The solution for 2y^{''}+8y=0,y^'(0)=1,y(0)=0 is y= 1/2 sin(2t)