What is the solution for y^{''}+4y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+4y=0,y(0)=1,y^'(0)=1 is y=cos(2t)+1/2 sin(2t)

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y^{''}+4y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 2

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

y^{^{'''}} + 4 y^{^{''}} - 9 y^{^{'}} - 36 y = 0

What is the solution for y^{''}+4y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+4y=0,y(0)=1,y^'(0)=1 is y=cos(2t)+1/2 sin(2t)