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

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

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

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

y = - 2 sin (t)

Solution steps

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

Solve linear ODE:

y = - 2 sin (t)

y = - 2 sin (t)

(D^{4} + D^{3} - 3 D^{2} - 5 D - 2) y = 0

x^{^{''}} + 5 x^{^{'}} + 6 x = 0, x (0) = 2, x^{^{'}} (0) = 3

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

10 y^{^{''}} + 9 y^{^{'}} + 2 y^{^{'}} = 0

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

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