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)

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = a

y = 3 x + 4, (7, 5)

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

x \to 1 lim (e^{x - 1} - 1)

f (x) = \frac{x ^{2}}{( 4 + x )}

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)