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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

3 y^{^{''}} + 5 y^{^{'}} + 2 y = t + e^{t}

3 y^{^{''}} + y^{^{'}} - 2 y = 2 cos (x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 t e^{4 t}

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

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

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