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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

e^{\frac{1}{x}}

ln (4 - x^{2})

x \to 2 lim (\frac{3}{∣ x - 2 ∣})

y d x + (- 2 x - 2 y) d y = 0

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

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