What is the solution for (D^2+9)y=cos(2x) ?

The solution for (D^2+9)y=cos(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/5 cos(2x)

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(D^2+9)y=cos(2x)

(D^{2} + 9) y = cos (2 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{5} cos (2 x)

Solution steps

(D^{2} + 9) y = cos (2 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{5} cos (2 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{5} cos (2 x)

y^{^{''}} + y = t, y (0) = 4, y (0) = 0, y^{^{'}} (0) = 0

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

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

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

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

What is the solution for (D^2+9)y=cos(2x) ?
The solution for (D^2+9)y=cos(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/5 cos(2x)