What is the solution for y^{''}+6y^'+5y=20 ?

The solution for y^{''}+6y^'+5y=20 is y=c_{1}e^{-t}+c_{2}e^{-5t}+4

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y^{''}+6y^'+5y=20

y^{^{''}} + 6 y^{^{'}} + 5 y = 20

y = c_{1} e^{- t} + c_{2} e^{- 5 t} + 4

Solution steps

y^{''} (t) + 6 y^{'} (t) + 5 y = 20

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 5 t} + 4

y = c_{1} e^{- t} + c_{2} e^{- 5 t} + 4

y^{^{''}} + y = e^{3 t} - 4 t cos (5 t)

y^{^{''''}} + 8 y^{^{''}} + 16 y = cos (x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 6 + x^{- 1} e^{x}

2 y^{^{''}} - 4 y^{^{'}} - 6 y = 3 e^{2 x}

y^{^{''}} + 2 y^{^{'}} = 2 x + 3 - e - 2 x

What is the solution for y^{''}+6y^'+5y=20 ?
The solution for y^{''}+6y^'+5y=20 is y=c_{1}e^{-t}+c_{2}e^{-5t}+4