What is the solution for y^{''}+16y=-24sin(4t) ?

The solution for y^{''}+16y=-24sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+3tcos(4t)

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y^{''}+16y=-24sin(4t)

y^{^{''}} + 16 y = - 24 sin (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + 3 t cos (4 t)

Solution steps

y^{''} (t) + 16 y = - 24 sin (4 t)

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + 3 t cos (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + 3 t cos (4 t)

\int x^{2} x^{3} + 2 d x

n = 0 \sum \infty \frac{1}{4 ^{n + 1}}

\frac{d}{d x} (e^{x^{2} - 2 x + 1})

\frac{\partial}{\partial x} (x^{3} y^{2} e^{x z})

\frac{d}{d x} (\frac{1 + e ^{- 2 x}}{2 + cos ( 2 x )})

What is the solution for y^{''}+16y=-24sin(4t) ?
The solution for y^{''}+16y=-24sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+3tcos(4t)