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

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

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

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

y = c_{1} e^{3 t} + c_{2} e^{- 4 t} - 1

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 4 t} - 1

y = c_{1} e^{3 t} + c_{2} e^{- 4 t} - 1

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 8 e x

y^{^{''}} - 36 y^{^{'}} = - 42 e^{- 5 t}, y (0) = 7, y^{^{'}} (0) = 0

2 \frac{d ^{2} y}{d t ^{2}} + 20 \frac{d y}{d t} + 50 = 8 \cdot (t)

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

x^{^{''}} - 3 x^{^{'}} + 2 x = 2 sin (t) - 6 cos (t)

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