What is the solution for y^{''}+4y=-16sin(2t) ?

The solution for y^{''}+4y=-16sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4tcos(2t)

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y^{''}+4y=-16sin(2t)

y^{^{''}} + 4 y = - 16 sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 t cos (2 t)

Solution steps

y^{''} (t) + 4 y = - 16 sin (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 t cos (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 t cos (2 t)

4 x^{2} (\frac{d y}{d x}) = (\frac{d y}{d x}) + 6 x e^{- y}

x \to \infty lim (4 - 5 x)

\int sin (9 π x) d x

\int (x^{3} e^{x^{2}}) d x

y^{^{''}} + 16 y = 4 csc^{2} (4 t)

What is the solution for y^{''}+4y=-16sin(2t) ?
The solution for y^{''}+4y=-16sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4tcos(2t)