What is the solution for y^{''}+4y=2sin(t) ?

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

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y^{''}+4y=2sin(t)

y^{^{''}} + 4 y = 2 sin (t)

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

Solution steps

y^{''} (t) + 4 y = 2 sin (t)

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 10 y = 54 e^{7 t}, y (0) = 5

y^{^{''}} + 9 y = - 2 e^{3} t

y^{^{'''}} - 2 y^{^{''}} = 4

y^{^{''}} - 9 y = e^{(- 2 x)} (- 17 + 10 x)

y^{^{''}} + 2 y^{^{'}} - 3 y = x^{2} sin (x)

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