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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{t} )}{( 1 + t ^{2} ) + 3 e ^{t}}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = x^{- 2} e^{x}

x^{^{''}} + 2 x^{^{'}} + 5 x = sin (2 t)

y^{^{''}} - 9 y^{^{'}} = 3 x^{2} - 5 sin (3 x)

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

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