What is the solution for y^{''}+9y=0,y(0)=3,y^'(0)=12 ?

The solution for y^{''}+9y=0,y(0)=3,y^'(0)=12 is y=3cos(3t)+4sin(3t)

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y^{''}+9y=0,y(0)=3,y^'(0)=12

y^{^{''}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 12

y = 3 cos (3 t) + 4 sin (3 t)

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

y = 3 cos (3 t) + 4 sin (3 t)

y = 3 cos (3 t) + 4 sin (3 t)

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

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 6 y^{^{'}} + 5 y = 0, y^{0} = 3

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

77 y^{^{''''}} + 204 y^{^{''}} + 64 y = 0

What is the solution for y^{''}+9y=0,y(0)=3,y^'(0)=12 ?
The solution for y^{''}+9y=0,y(0)=3,y^'(0)=12 is y=3cos(3t)+4sin(3t)