What is the solution for y^{''}+4y=0,y(0)=2,y^'(0)=-1 ?

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

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y^{''}+4y=0,y(0)=2,y^'(0)=-1

y^{^{''}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 5 y = 0, y (0) = 0

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

y^{^{''''}} - 9 y^{^{'''}} + 19 y^{^{''}} + 29 y^{^{'}} = 0

y^{^{'''}} + 21 y^{^{''}} + 117 y^{^{'}} + 97 y = 0

36 y^{^{''}} - 60 y^{^{'}} + 25 y = 0

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