What is the solution for y^{''}+4y^'+3y=6 ?

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

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y^{''}+4y^'+3y=6

y^{^{''}} + 4 y^{^{'}} + 3 y = 6

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 6

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 1, 6 t y (0) = 3, y^{^{'}} (0) = - 6

y^{^{''}} + 2 y^{^{'}} + 2 y = a cos (b t)

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (4 t)

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{13 e ^{t}}{1 + t ^{2}}

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