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

The solution for y^{''}+2y^'-3y=e^{2t}-2t is y=c_{1}e^t+c_{2}e^{-3t}+1/5 e^{2t}+(2t)/3+4/9

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

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

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{1}{5} e^{2 t} + \frac{2 t}{3} + \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{1}{5} e^{2 t} + \frac{2 t}{3} + \frac{4}{9}

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{1}{5} e^{2 t} + \frac{2 t}{3} + \frac{4}{9}

x^{^{''}} + 36 x = 8 cos (6 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 7 x + 28

4 y^{^{''}} - y = e^{x}

y^{^{''}} - 2 y^{^{'}} + 1 = x^{2}

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 t e^{t}, y (0) = 1, y^{^{'}} (0) = - 5

What is the solution for y^{''}+2y^'-3y=e^{2t}-2t ?
The solution for y^{''}+2y^'-3y=e^{2t}-2t is y=c_{1}e^t+c_{2}e^{-3t}+1/5 e^{2t}+(2t)/3+4/9