What is the solution for y^{''}+4y^'+8y=160 ?

The solution for y^{''}+4y^'+8y=160 is y=e^{-2t}(c_{1}cos(2t)+c_{2}sin(2t))+20

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y^{''}+4y^'+8y=160

y^{^{''}} + 4 y^{^{'}} + 8 y = 160

y = e^{- 2 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 20

Solution steps

y^{''} (t) + 4 y^{'} (t) + 8 y = 160

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 20

y = e^{- 2 t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + 20

- x^{^{''}} + 4 x + 4 = 0

y^{^{''}} - 6 y^{^{'}} + 45 y = 2 sin (3 t)

\frac{d ^{2} y}{d x ^{2}} (1) + \frac{d y}{d x} (1) = tan (x)

y^{^{''}} + 4 y = sin (x) \cdot sin (2 x)

y^{^{'}} + 2 y^{^{''}} - 4 t e^{t} = 9, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+8y=160 ?
The solution for y^{''}+4y^'+8y=160 is y=e^{-2t}(c_{1}cos(2t)+c_{2}sin(2t))+20