What is the solution for y^{''}+4y=3cos(at) ?

The solution for y^{''}+4y=3cos(at) is y=c_{1}cos(2t)+c_{2}sin(2t)+(3cos(at))/(-a^2+4)

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y^{''}+4y=3cos(at)

y^{^{''}} + 4 y = 3 cos (a t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 cos ( a t )}{- a ^{2} + 4}

Solution steps

y^{''} (t) + 4 y = 3 cos (a t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 cos ( a t )}{- a ^{2} + 4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 cos ( a t )}{- a ^{2} + 4}

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 7 e^{- 3 x}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 2 x^{2} - x + 1

y^{^{''}} - 3 y - 5 = e^{2} - 3 cos (3 x)

y^{^{''}} + 36 y = 24 sin (6 x)

y^{^{''}} + 6 y = g (t)

What is the solution for y^{''}+4y=3cos(at) ?
The solution for y^{''}+4y=3cos(at) is y=c_{1}cos(2t)+c_{2}sin(2t)+(3cos(at))/(-a^2+4)