What is the solution for y^{''}-y^'-6y=2 ?

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

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y^{''}-y^'-6y=2

y^{^{''}} - y^{^{'}} - 6 y = 2

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

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 2

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} - 1 = 0

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 y = 10 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} + 9 y = tan (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 96 e^{- 5 t} - 24 e^{- 3 t} + 18

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