What is the solution for y^{''}-4y^'-12y=4 ?

The solution for y^{''}-4y^'-12y=4 is y=c_{1}e^{6t}+c_{2}e^{-2t}-1/3

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y^{''}-4y^'-12y=4

y^{^{''}} - 4 y^{^{'}} - 12 y = 4

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

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = 4

Solve linear ODE:

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

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 3 x^{2} + 5 x

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 0, y^{^{'}} (0) = a

y^{^{''}} + 4 y = sin^{22} (x)

y^{^{''}} - 3 y^{^{'}} = 2 - 6 x

What is the solution for y^{''}-4y^'-12y=4 ?
The solution for y^{''}-4y^'-12y=4 is y=c_{1}e^{6t}+c_{2}e^{-2t}-1/3