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

The solution for y^{''}+y^'-6y=2t is y=c_{1}e^{2t}+c_{2}e^{-3t}-t/3-1/18

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y^{''}+y^'-6y=2t

y^{^{''}} + y^{^{'}} - 6 y = 2 t

y = c_{1} e^{2 t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{1}{18}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{1}{18}

y = c_{1} e^{2 t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{1}{18}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 8 + 2 sin (x)

y^{^{''}} + y = 3 sin (2 t) + t cos (2 t)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 2 y = 2 x

y^{^{''}} - 2 y^{^{'}} + y = \frac{11 e ^{t}}{1 + t ^{2}}

(D^{2} + 4) y = sin (2 x)

What is the solution for y^{''}+y^'-6y=2t ?
The solution for y^{''}+y^'-6y=2t is y=c_{1}e^{2t}+c_{2}e^{-3t}-t/3-1/18