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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = sec^{3} (x) - sec (x)

y^{^{''''}} - 5 y^{^{''}} + 4 y = 8 cos (2 x) - 12 e^{- x}

x^{^{''}} + x = 9 cos (8 t)

y^{^{''}} + y = - sec (x)

(D^{2} - 1) y = e^{x}

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