What is the solution for 3x^{''}+12x=sin(2t) ?

The solution for 3x^{''}+12x=sin(2t) is x=c_{1}cos(2t)+c_{2}sin(2t)-(tcos(2t))/(12)

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3x^{''}+12x=sin(2t)

3 x^{^{''}} + 12 x = sin (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t cos ( 2 t )}{12}

Solution steps

3 x^{''} (t) + 12 x = sin (2 t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t cos ( 2 t )}{12}

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{t cos ( 2 t )}{12}

y^{^{''}} + 25 y = 2 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} + 2 y^{^{''}} = 12 x + 4

y^{^{''}} + 25 y = 25 sec^{2} (5 t)

x^{''} + 2 x = 8, x (π) = 0, x (0) = 0

y^{^{''}} - y^{^{'}} - 42 y = - 122 t + 84 t^{2}

What is the solution for 3x^{''}+12x=sin(2t) ?
The solution for 3x^{''}+12x=sin(2t) is x=c_{1}cos(2t)+c_{2}sin(2t)-(tcos(2t))/(12)