What is the solution for y^{''}+5y^'+4y=20 ?

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

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y^{''}+5y^'+4y=20

y^{^{''}} + 5 y^{^{'}} + 4 y = 20

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

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = 20

Solve linear ODE:

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

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

y^{^{''}} + 4 y = cos (\frac{9}{4} t)

\frac{d ^{4} y}{d x ^{4}} + \frac{d ^{2} y}{d x ^{2}} = cos (4 x)

y^{^{''''}} - 3 y^{^{''}} - 4 y = 10 sin (x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 25 y = 8 e^{5 x}

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