What is the solution for 1/16 (d^2x)/(dt^2)+x=10sin(4t) ?

The solution for 1/16 (d^2x)/(dt^2)+x=10sin(4t) is x=c_{1}cos(4t)+c_{2}sin(4t)-20tcos(4t)

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1/16 (d^2x)/(dt^2)+x=10sin(4t)

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} + x = 10 sin (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) - 20 t cos (4 t)

Solution steps

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} + x = 10 sin (4 t)

Solve linear ODE:

x = c_{1} cos (4 t) + c_{2} sin (4 t) - 20 t cos (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) - 20 t cos (4 t)

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

y^{^{''}} + 10 y^{^{'}} + 24 y = 2 e^{- 4 x} (4 x - 2)

1 y^{^{''}} + 25 y = cos (5 x) csc^{2} (5 x)

y^{^{''}} + 8 y^{^{'}} + 16 y = 2 sin (4 t)

y^{^{''}} + 4 y = 14 cos (2 x) + 21 sin (2 x)

What is the solution for 1/16 (d^2x)/(dt^2)+x=10sin(4t) ?
The solution for 1/16 (d^2x)/(dt^2)+x=10sin(4t) is x=c_{1}cos(4t)+c_{2}sin(4t)-20tcos(4t)