What is the solution for y^{''}+7y^'+6y=8 ?

The solution for y^{''}+7y^'+6y=8 is y=c_{1}e^{-t}+c_{2}e^{-6t}+4/3

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y^{''}+7y^'+6y=8

y^{^{''}} + 7 y^{^{'}} + 6 y = 8

y = c_{1} e^{- t} + c_{2} e^{- 6 t} + \frac{4}{3}

Solution steps

y^{''} (t) + 7 y^{'} (t) + 6 y = 8

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 6 t} + \frac{4}{3}

y = c_{1} e^{- t} + c_{2} e^{- 6 t} + \frac{4}{3}

y^{^{''}} + 4 y = 3 sin (t), y (0) = 2, y^{^{'}} (0) = - 1

y^{^{''}} - 4 y = 3 e^{2 x}, y (0) = 0, y^{^{'}} (0) = 0

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

x^{^{''}} + 0.8 x^{^{'}} + 19.6 x = 4 e^{- 2 t} cos (6 t)

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+7y^'+6y=8 ?
The solution for y^{''}+7y^'+6y=8 is y=c_{1}e^{-t}+c_{2}e^{-6t}+4/3