What is the solution for y^{''}+a^2y=cos(bt) ?

The solution for y^{''}+a^2y=cos(bt) is y=c_{1}cos(at)+c_{2}sin(at)+(cos(bt))/(a^2-b^2)

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y^{''}+a^2y=cos(bt)

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

y = c_{1} cos (a t) + c_{2} sin (a t) + \frac{cos ( b t )}{a ^{2} - b ^{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (a t) + c_{2} sin (a t) + \frac{cos ( b t )}{a ^{2} - b ^{2}}

y = c_{1} cos (a t) + c_{2} sin (a t) + \frac{cos ( b t )}{a ^{2} - b ^{2}}

y^{^{''}} - 12 y^{^{'}} + 36 y = 7 + 2 x

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

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

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

y^{^{''}} + y^{^{'}} = cos (3 t), y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+a^2y=cos(bt) ?
The solution for y^{''}+a^2y=cos(bt) is y=c_{1}cos(at)+c_{2}sin(at)+(cos(bt))/(a^2-b^2)