What is the solution for y^{''}+4y^'+20y=80 ?

The solution for y^{''}+4y^'+20y=80 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))+4

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y^{''}+4y^'+20y=80

y^{^{''}} + 4 y^{^{'}} + 20 y = 80

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 4

Solution steps

y^{''} (t) + 4 y^{'} (t) + 20 y = 80

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 4

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 4

y = x, y = x - 2, y = 0

\int \frac{1}{( x ^{2} + 81 ) ^{2}} d x

\int 4 cos^{2} (θ) d θ

\int_{0}^{\frac{1}{2}} tan (\frac{x}{2}) d x

x \to 0 lim (\frac{2 e ^{x} - 2}{x})

What is the solution for y^{''}+4y^'+20y=80 ?
The solution for y^{''}+4y^'+20y=80 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))+4