What is the solution for y^{''}-y^'=8-(t-2) ?

The solution for y^{''}-y^'=8-(t-2) is y=c_{1}e^t+c_{2}+(t^2)/2-9t

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 16 x^{^{'}} + 64 x = 5 t e^{8 t}

y^{^{''}} - 5 y^{^{'}} + 6 y = - 2 x + 1

0.05 y^{^{''}} + 5 y^{^{'}} + \frac{1}{0.006} y = 5 sin (25 x)

(d^{2} - d) y = e^{x} cos (x)

y^{^{''}} + 17 y^{^{'}} + 30 y = cos (2 t)

What is the solution for y^{''}-y^'=8-(t-2) ?
The solution for y^{''}-y^'=8-(t-2) is y=c_{1}e^t+c_{2}+(t^2)/2-9t