What is the solution for y^{''}+12y^'+100y=48sin(10x) ?

The solution for y^{''}+12y^'+100y=48sin(10x) is y=e^{-6x}(c_{1}cos(8x)+c_{2}sin(8x))-2/5 cos(10x)

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y^{''}+12y^'+100y=48sin(10x)

y^{^{''}} + 12 y^{^{'}} + 100 y = 48 sin (10 x)

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

Solution steps

y^{''} (x) + 12 y^{'} (x) + 100 y = 48 sin (10 x)

Solve linear ODE:

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

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

y^{^{''}} - 16 y = 7 t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} + 2 y - x - 3 = 0, y (0) = 0

\frac{d ^{2} x}{d t ^{2}} - 3 \frac{d x}{d t} + 2 x = e^{t}

y^{^{''}} - y = sin (3 x)

5 y^{^{''}} - 19 y^{^{'}} - 4 y = 1 + 4 x

What is the solution for y^{''}+12y^'+100y=48sin(10x) ?
The solution for y^{''}+12y^'+100y=48sin(10x) is y=e^{-6x}(c_{1}cos(8x)+c_{2}sin(8x))-2/5 cos(10x)