What is the solution for y^{''}+16y=8sin(4t) ?

The solution for y^{''}+16y=8sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)-tcos(4t)

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y^{''}+16y=8sin(4t)

y^{^{''}} + 16 y = 8 sin (4 t)

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

Solution steps

y^{''} (t) + 16 y = 8 sin (4 t)

Solve linear ODE:

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

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

\int \frac{1}{x ^{2} + 4 x + 9} d x

y (t + 5) (t + 2) y^{^{'}} + t + 3 = 0

y = (x - 10)^{- 3}

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

\int \frac{1}{( x - 1 ) ( x ^{2} + 3 )} d x

What is the solution for y^{''}+16y=8sin(4t) ?
The solution for y^{''}+16y=8sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)-tcos(4t)