What is the solution for y^{''}+4y=cos(9/4 t) ?

The solution for y^{''}+4y=cos(9/4 t) is y=c_{1}cos(2t)+c_{2}sin(2t)-16/17 cos((9t)/4)

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y^{''}+4y=cos(9/4 t)

y^{^{''}} + 4 y = cos (\frac{9}{4} t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{16}{17} cos (\frac{9 t}{4})

Solution steps

y^{''} (t) + 4 y = cos (\frac{9}{4} t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{16}{17} cos (\frac{9 t}{4})

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{16}{17} cos (\frac{9 t}{4})

\frac{d ^{4} y}{d x ^{4}} + \frac{d ^{2} y}{d x ^{2}} = cos (4 x)

y^{^{''''}} - 3 y^{^{''}} - 4 y = 10 sin (x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 25 y = 8 e^{5 x}

y^{^{''}} + 4 y = t^{2} + 2 e^{t}

What is the solution for y^{''}+4y=cos(9/4 t) ?
The solution for y^{''}+4y=cos(9/4 t) is y=c_{1}cos(2t)+c_{2}sin(2t)-16/17 cos((9t)/4)