What is the solution for y^{''}+9y=cos(2t),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+9y=cos(2t),y(0)=1,y^'(0)=0 is y= 4/5 cos(3t)+1/5 cos(2t)

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y^{''}+9y=cos(2t),y(0)=1,y^'(0)=0

y^{^{''}} + 9 y = cos (2 t), y (0) = 1, y^{^{'}} (0) = 0

y = \frac{4}{5} cos (3 t) + \frac{1}{5} cos (2 t)

Solution steps

y^{''} (t) + 9 y = cos (2 t)

Solve linear ODE:

y = \frac{4}{5} cos (3 t) + \frac{1}{5} cos (2 t)

y = \frac{4}{5} cos (3 t) + \frac{1}{5} cos (2 t)

4 y^{^{''}} + 324 y = 8 sin (9 t)

y^{^{''}} + y = 1 sin (18.8 t)

D (x) = 0.2 x^{2} + 9.2 x - 1.3

y^{^{''}} + 5 y^{^{'}} = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 x + 5, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{''}+9y=cos(2t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+9y=cos(2t),y(0)=1,y^'(0)=0 is y= 4/5 cos(3t)+1/5 cos(2t)