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)

y^{^{''}} - y = 0, y (- 1) = 5, y^{^{'}} (- 1) = - 5

y^{^{''}} - y^{^{'}} = 0, y^{^{'}} (0) = 0

9 y^{^{''}} - 12 y + 4 y = 0

y^{^{''''}} - 16 y^{^{'''}} + 64 y^{^{''}} = 0

2 y^{^{''}} + 36 y^{^{'}} + 163 y = 0, y (0) = 2, y^{^{'}} (0) = 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)