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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y = e^{x} cos (x)

y^{^{''}} + 4 y^{^{'}} - 2 y = 8 sin (2 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t + 12 e^{- t}

y^{^{''}} - 64 y = cosh (8 x)

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 6 y = x e^{2 x}

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