What is the solution for (D^2+1)y=3sin(2t)-2cos(2t) ?

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

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(D^2+1)y=3sin(2t)-2cos(2t)

(D^{2} + 1) y = 3 sin (2 t) - 2 cos (2 t)

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

Solution steps

(D^{2} + 1) y = 3 sin (2 t) - 2 cos (2 t)

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} = \frac{1}{cos ( x )}

y^{^{'''}} - 4 y^{^{'}} = e^{x} + 4 x

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

y^{^{''''}} - 5 y^{^{''}} + 4 y = e^{x} (- 122 - 48 x + 72 x^{2})

y^{^{''}} + y^{^{'}} - 12 y = e^{t} + e^{2 t} - 1

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