What is the solution for y^{''}-4y^'-12y=12t,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}-4y^'-12y=12t,y(0)=1,y^'(0)=-1 is y=(-1/2+2/3)e^{6t}+1/2 e^{-2t}-t+1/3

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y^{''}-4y^'-12y=12t,y(0)=1,y^'(0)=-1

y^{^{''}} - 4 y^{^{'}} - 12 y = 12 t, y (0) = 1, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} + 2 y^{^{''}} + y = 5 + cos (6 t)

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

y^{^{''}} + 36 y = 5 sec (6 x)

y^{^{''}} - 4 y^{^{'}} = (t^{2} + 1) sin (2 t)

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t)

What is the solution for y^{''}-4y^'-12y=12t,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-4y^'-12y=12t,y(0)=1,y^'(0)=-1 is y=(-1/2+2/3)e^{6t}+1/2 e^{-2t}-t+1/3