What is the solution for y^{''}+y^'-6y=18 ?

The solution for y^{''}+y^'-6y=18 is y=c_{1}e^{2t}+c_{2}e^{-3t}-3

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y^{''}+y^'-6y=18

y^{^{''}} + y^{^{'}} - 6 y = 18

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

Solution steps

y^{''} (t) + y^{'} (t) - 6 y = 18

Solve linear ODE:

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

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

x^{^{''}} + 4 x^{^{'}} + 20 x = 4 cos (2 t)

y^{^{''}} + y^{^{'}} - 6 y = - 5 e^{- 2 x}

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

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{4 x}

x^{^{''}} + x = \frac{1}{cos ( t )}

What is the solution for y^{''}+y^'-6y=18 ?
The solution for y^{''}+y^'-6y=18 is y=c_{1}e^{2t}+c_{2}e^{-3t}-3