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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 100 y = cosh (10 x)

y^{^{''}} - y^{^{'}} + 5 y = e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = 8 e^{x} - 5

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = 16 sin (x)

y^{^{''}} - 2 y^{^{'}} = 7 cos (3 t)

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