What is the solution for 3y^{''}+12y=-4sin(2t) ?

The solution for 3y^{''}+12y=-4sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tcos(2t))/3

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3y^{''}+12y=-4sin(2t)

3 y^{^{''}} + 12 y = - 4 sin (2 t)

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

Solution steps

3 y^{''} (t) + 12 y = - 4 sin (2 t)

Solve linear ODE:

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

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

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

y^{^{''}} - 8 y^{^{'}} + 15 y = 225 x

y^{^{''}} - 3 y = e^{x}

y^{^{''}} + y^{^{'}} + y = sinh (x)

\frac{d ^{2} y}{d t ^{2}} - 4 y = t cos (2 t), y = 0, t = 0

What is the solution for 3y^{''}+12y=-4sin(2t) ?
The solution for 3y^{''}+12y=-4sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tcos(2t))/3