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

The solution for y^{''}+16y=2sin(2t) is y=c_{1}cos(4t)+c_{2}sin(4t)+1/6 sin(2t)

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

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

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{6} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{6} sin (2 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{6} sin (2 t)

\frac{d}{d x} (x^{2} - 5)

\int - 13 cos^{3} (x) sin (x) d x

\int_{0}^{1} \frac{5}{x ( x + 1 )} d x

(2 x y + 2 x^{3} y) d x - (1 + x^{2}) d y = 0

x \to \infty lim (a x^{4} + b x^{2} + c)

What is the solution for y^{''}+16y=2sin(2t) ?
The solution for y^{''}+16y=2sin(2t) is y=c_{1}cos(4t)+c_{2}sin(4t)+1/6 sin(2t)