What is the solution for y^{''}+3y^'+5=0 ?

The solution for y^{''}+3y^'+5=0 is y=c_{1}+c_{2}e^{-3t}-(5t)/3

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y^{''}+3y^'+5=0

y^{^{''}} + 3 y^{^{'}} + 5 = 0

y = c_{1} + c_{2} e^{- 3 t} - \frac{5 t}{3}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 5 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} - \frac{5 t}{3}

y = c_{1} + c_{2} e^{- 3 t} - \frac{5 t}{3}

y^{^{''}} - 5 y^{^{'}} - 24 y = 5 + 2 t

y^{^{''}} - y^{^{'}} - 2 y = 3 t^{2} - 1

y^{^{''}} + 36 y = 33 cos (5 x)

y^{^{'''}} - y^{^{'}} = 2 \cdot cos (t)

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

What is the solution for y^{''}+3y^'+5=0 ?
The solution for y^{''}+3y^'+5=0 is y=c_{1}+c_{2}e^{-3t}-(5t)/3