What is the solution for y^{''}-4y^'-12y=3e^5t ?

The solution for y^{''}-4y^'-12y=3e^5t is y=c_{1}e^{6t}+c_{2}e^{-2t}-(e^5t)/4+(e^5)/(12)

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y^{''}-4y^'-12y=3e^5t

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 e^{5} t

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{5} t}{4} + \frac{e ^{5}}{12}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = 3 e^{5} t

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{5} t}{4} + \frac{e ^{5}}{12}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{5} t}{4} + \frac{e ^{5}}{12}

y^{^{'''}} - 4 y^{^{''}} - y^{^{'}} + 4 y = e^{6 x}

y^{^{''}} + 36 y = 2 cos (6 x) + 3 sin (6 x)

y^{^{''}} - 3 y^{^{'}} = t e^{3 t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 12 y^{^{'}} + 35 y = - 40 e^{-} 3 t

y^{^{''}} + 2 y^{^{'}} + 5 y = g (t), y (0) = - 4, y^{^{'}} (0) = 0

What is the solution for y^{''}-4y^'-12y=3e^5t ?
The solution for y^{''}-4y^'-12y=3e^5t is y=c_{1}e^{6t}+c_{2}e^{-2t}-(e^5t)/4+(e^5)/(12)