What is the solution for y^{''}+4y^'+5=0 ?

The solution for y^{''}+4y^'+5=0 is y=c_{1}+c_{2}e^{-4t}-(5t)/4

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y^{''}+4y^'+5=0

y^{^{''}} + 4 y^{^{'}} + 5 = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 = 0

Solve linear ODE:

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

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

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

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 2 t

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (x) + 1

y^{^{''}} + 36 y = - 12 sin (6 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 6 e^{- t}

What is the solution for y^{''}+4y^'+5=0 ?
The solution for y^{''}+4y^'+5=0 is y=c_{1}+c_{2}e^{-4t}-(5t)/4