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

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

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

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

y = \frac{1}{3} cos (2 t) - \frac{1}{3} cos (t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{3} cos (2 t) - \frac{1}{3} cos (t)

y = \frac{1}{3} cos (2 t) - \frac{1}{3} cos (t)

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