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)

y^{^{''}} + 2 y^{^{'}} + 5 y = 50 x sin (x) + 100 x cos (x)

y^{^{''}} - 4 y = - 7 e^{2 x} + x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 3 y = e^{2 x} sin (x)

y^{^{''}} + 9 y = e^{2 x}

y^{^{''}} + y = 3 + e^{2 x}

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)