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)

D^{2} (D - 2)^{3} y = 48 e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}

y^{^{''}} - 3 y^{^{'}} - 40 y = 210 e^{2 t}

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = t^{2} - 2 t

y^{^{''}} - 2 y^{^{'}} + 6 y = 18 sin (3 t)

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)