What is the solution for x^{''}+4x=sin(3t) ?

The solution for x^{''}+4x=sin(3t) is x=c_{1}cos(2t)+c_{2}sin(2t)-1/5 sin(3t)

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x^{''}+4x=sin(3t)

x^{^{''}} + 4 x = sin (3 t)

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

Solution steps

x^{''} (t) + 4 x = sin (3 t)

Solve linear ODE:

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

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

(D^{2} - 4 D + 3) y = 9 x^{2} - 36 x + 37

y^{^{''}} + 4 y = x^{2} e^{- x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} - 3 y = 6, y (0) = 9, y^{^{'}} (0) = 27

y^{^{''}} + y^{^{'}} - 6 y = 3 x + 2

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x + 2

What is the solution for x^{''}+4x=sin(3t) ?
The solution for x^{''}+4x=sin(3t) is x=c_{1}cos(2t)+c_{2}sin(2t)-1/5 sin(3t)