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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 14 y^{^{'}} + 40 y = 0, y (0) = 9, y^{^{'}} (0) = - 48

y^{^{'''}} - 7 y^{^{''}} + 6 y = 0

2 y^{^{''}} - y^{^{'}} = 0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

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