What is the solution for y^{''}+2y^'+5=0 ?

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

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y^{''}+2y^'+5=0

y^{^{''}} + 2 y^{^{'}} + 5 = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 = 0

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 4 y = 5 x^{4} + 3 x^{2} - x

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

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

4 y^{^{''}} + 36 y = sec (3 x)

y^{^{''}} - 4 y^{^{'}} - 5 y = 24 x e^{5 x} - 26 sin (x)

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