What is the solution for y^{''}+5y^'+6y=3 ?

The solution for y^{''}+5y^'+6y=3 is y=c_{1}e^{-2t}+c_{2}e^{-3t}+1/2

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y^{''}+5y^'+6y=3

y^{^{''}} + 5 y^{^{'}} + 6 y = 3

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

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = 3

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 25 y = (5 x^{3} - 7) e^{- x}

x^{^{''}} + 4 x = cos (t)

y^{^{''}} + 9 y = cos (8 x)

(d^{2} - 2 d + 1) y = x^{2} e^{2} x

y^{^{''}} - 6 y^{^{'}} + 9 y = 18 x + 5

What is the solution for y^{''}+5y^'+6y=3 ?
The solution for y^{''}+5y^'+6y=3 is y=c_{1}e^{-2t}+c_{2}e^{-3t}+1/2