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

The solution for -3y^{''}-2y^'+y=t^2+t-2e^{2t} is y=c_{1}e^{-t}+c_{2}e^{t/3}+t^2+5t+16+2/15 e^{2t}

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

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

y = c_{1} e^{- t} + c_{2} e^{\frac{t}{3}} + t^{2} + 5 t + 16 + \frac{2}{15} e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{\frac{t}{3}} + t^{2} + 5 t + 16 + \frac{2}{15} e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{\frac{t}{3}} + t^{2} + 5 t + 16 + \frac{2}{15} e^{2 t}

x^{^{'''}} - 4 x^{^{''}} + 5 x^{^{'}} - 2 x = 2 t + 3

y^{^{''}} + 4 y = sin (2 x + 3)

y^{^{''}} - 2 y^{^{'}} + y = tan (x)

y^{^{''}} + y^{^{'}} - 12 y = - 40 t + 24 t^{2}

y^{^{''}} + y^{^{'}} + 4 y = e^{t} + e^{- t}

What is the solution for -3y^{''}-2y^'+y=t^2+t-2e^{2t} ?
The solution for -3y^{''}-2y^'+y=t^2+t-2e^{2t} is y=c_{1}e^{-t}+c_{2}e^{t/3}+t^2+5t+16+2/15 e^{2t}