What is the solution for y^{''}+12y^'+100y=48sin(10t) ?

The solution for y^{''}+12y^'+100y=48sin(10t) is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))-2/5 cos(10t)

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y^{''}+12y^'+100y=48sin(10t)

y^{^{''}} + 12 y^{^{'}} + 100 y = 48 sin (10 t)

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) - \frac{2}{5} cos (10 t)

Solution steps

y^{''} (t) + 12 y^{'} (t) + 100 y = 48 sin (10 t)

Solve linear ODE:

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) - \frac{2}{5} cos (10 t)

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) - \frac{2}{5} cos (10 t)

x^{^{''}} + 25 x = 102 cos (t)

y^{^{''}} - y^{^{'}} = 3 - 4 x

y^{^{''}} + 3 y + 2 y = cos (e^{x})

y^{^{''}} + 15 y^{^{'}} + 7 y = sin (x) + cos (x)

(y^{^{''}}) - 2 y^{^{'}} + y = \frac{e ^{x}}{1 + x ^{2}}

What is the solution for y^{''}+12y^'+100y=48sin(10t) ?
The solution for y^{''}+12y^'+100y=48sin(10t) is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))-2/5 cos(10t)