What is the solution for y^{''}+4y^'+5y=8 ?

The solution for y^{''}+4y^'+5y=8 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+8/5

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y^{''}+4y^'+5y=8

y^{^{''}} + 4 y^{^{'}} + 5 y = 8

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 8

Solve linear ODE:

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

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

- 5 x^{2} + 4

\frac{d}{d x} (6 - 5 cos (\frac{π x}{15}))

\int \frac{x ^{4} + 4}{x ^{4}} d x

\frac{d}{d x} (x - 1 + x + 1)

What is the solution for y^{''}+4y^'+5y=8 ?
The solution for y^{''}+4y^'+5y=8 is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+8/5