What is the solution for y^{''}+4y=0,y(0)=6,y^{''}(0)=-4 ?

The solution for y^{''}+4y=0,y(0)=6,y^{''}(0)=-4 is y=6cos(2t)+c_{2}sin(2t)

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y^{''}+4y=0,y(0)=6,y^{''}(0)=-4

y^{^{''}} + 4 y = 0, y (0) = 6, y^{^{''}} (0) = - 4

y = 6 cos (2 t) + c_{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 6 cos (2 t) + c_{2} sin (2 t)

y = 6 cos (2 t) + c_{2} sin (2 t)

4 y^{^{''}} - 12 y^{^{'}} + 73 y = 0

81 y^{^{''}} - 4 y = 0

y^{^{'''}} + 3 y^{^{''}} - 7 y^{^{'}} + 3 y = 0

y^{^{''}} + (e - 1) y^{^{'}} - ey = 0

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

What is the solution for y^{''}+4y=0,y(0)=6,y^{''}(0)=-4 ?
The solution for y^{''}+4y=0,y(0)=6,y^{''}(0)=-4 is y=6cos(2t)+c_{2}sin(2t)