What is the solution for y^{''}-3y^'+2y=4t-8 ?

The solution for y^{''}-3y^'+2y=4t-8 is y=c_{1}e^{2t}+c_{2}e^t+2t-1

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y^{''}-3y^'+2y=4t-8

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t - 8

y = c_{1} e^{2 t} + c_{2} e^{t} + 2 t - 1

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 4 t - 8

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + 2 t - 1

y = c_{1} e^{2 t} + c_{2} e^{t} + 2 t - 1

(d^{3} - 2 d + 4) y = 3 x^{2} - 5 x + 2

(d^{3} - 3 d^{2} + 4 d - 2) \cdot y = e^{x} + cos (x) + x

(d^{4} + d^{2} + 16) y = 16 x^{2} + 256

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 5} x

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{x}

What is the solution for y^{''}-3y^'+2y=4t-8 ?
The solution for y^{''}-3y^'+2y=4t-8 is y=c_{1}e^{2t}+c_{2}e^t+2t-1