What is the solution for y^{''}+4y^'+3y=18 ?

The solution for y^{''}+4y^'+3y=18 is y=c_{1}e^{-t}+c_{2}e^{-3t}+6

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y^{''}+4y^'+3y=18

y^{^{''}} + 4 y^{^{'}} + 3 y = 18

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 18

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + y = - 13 sin (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 2 e^{4 t}, y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} - 5 y^{^{'}} + 6 y = sin (4 x)

y^{^{''}} - 2 y^{^{'}} + 4 y = e^{4 x}

What is the solution for y^{''}+4y^'+3y=18 ?
The solution for y^{''}+4y^'+3y=18 is y=c_{1}e^{-t}+c_{2}e^{-3t}+6