What is the solution for y^{''}+16y=40sin(4t) ?

The solution for y^{''}+16y=40sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)-5tcos(4t)

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y^{''}+16y=40sin(4t)

y^{^{''}} + 16 y = 40 sin (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) - 5 t cos (4 t)

Solution steps

y^{''} (t) + 16 y = 40 sin (4 t)

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) - 5 t cos (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) - 5 t cos (4 t)

\frac{\partial}{\partial y} (e^{x} + e^{y})

y^{^{''}} - 2 y^{^{'}} - y = 0, y (0) = 0, y^{^{'}} (0) = 22

1

y = e^{4 x}, at x = \frac{1}{4} ln (5)

\frac{d}{d x} ((4 + 8 x) cosh (2 x))

What is the solution for y^{''}+16y=40sin(4t) ?
The solution for y^{''}+16y=40sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)-5tcos(4t)