What is the solution for linear y^{''}+4y=0,y(0)=2,y^'(0)=2 ?

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

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linear y^{''}+4y=0,y(0)=2,y^'(0)=2

linear

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

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

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

49 y^{^{''}} - 126 y^{^{'}} + 117 y = 0, y (4) = 2, y^{^{'}} (4) = 2

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = 6 - 6 e^{3}, y (1) = 0

2 y^{^{''}} - 10 y^{^{'}} + 12 y = 0

y^{^{'''}} - 3 y^{^{''}} - 49 y^{^{'}} - 45 y = 0

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

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