What is the solution for y^{''}-4y^'+3y=et-2t+3 ?

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

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y^{''}-4y^'+3y=et-2t+3

y^{^{''}} - 4 y^{^{'}} + 3 y = e t - 2 t + 3

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 3 y = e t - 2 t + 3

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} = 2 x + 5 - e^{3 x}

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{3 t}

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

y^{^{''}} + y = 2 sin (2 t), y (0) = 11, y^{^{'}} (0) = 0

y^{^{''}} + y = e^{- 2 t} sin (t) + t \cdot cos (t)

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