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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

x^{^{''}} + 2 x^{^{'}} + x = 0, x (0) = 0, x^{^{'}} (0) = 1

x^{^{''}} + 28 x^{^{'}} + 196 x = 0

y^{^{''}} - 14 y^{^{'}} + 49 y = 0, y (0) = 1, y (1) = 0

y^{^{''}} + 3 y^{^{'}} = - 2 y

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