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

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

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y^{''}+y^'-6y=12e^{3t}+12e^{-2t}

y^{^{''}} + y^{^{'}} - 6 y = 12 e^{3 t} + 12 e^{- 2 t}

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

Solution steps

y^{''} (t) + y^{'} (t) - 6 y = 12 e^{3 t} + 12 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} = 2 x^{4}

y^{^{''}} + 5 y^{^{'}} + 2 y = e^{- x}

y^{^{''}} + 10 y^{^{'}} + 16 y = 2581 cos (5 t)

y^{^{''}} = k y - c

y^{^{''}} + 2 y^{^{'}} = sec (x)

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