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

The solution for y^{''}+2y^'-3y=t is y=c_{1}e^t+c_{2}e^{-3t}-t/3-2/9

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y^{''}+2y^'-3y=t

y^{^{''}} + 2 y^{^{'}} - 3 y = t

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{2}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{2}{9}

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{t}{3} - \frac{2}{9}

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{2} e^{3 x} - 8 e^{3 x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 10

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

y^{^{''}} - 2 y^{^{'}} + 1 = e^{t} arctan (t)

y^{^{''}} - y^{^{'}} - 6 y = 17 cos (3 x)

What is the solution for y^{''}+2y^'-3y=t ?
The solution for y^{''}+2y^'-3y=t is y=c_{1}e^t+c_{2}e^{-3t}-t/3-2/9