What is the solution for y^{''}-y^'=2t^2+t-1 ?

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

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y^{''}-y^'=2t^2+t-1

y^{^{''}} - y^{^{'}} = 2 t^{2} + t - 1

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

Solution steps

y^{''} (t) - y^{'} (t) = 2 t^{2} + t - 1

Solve linear ODE:

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

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

y^{^{''}} - 2 y = 12 x - 10

y^{^{''}} + 2 y^{^{'}} + y = 3 (t - 1) e^{- 1}

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

y^{^{''}} + y - 20 y = (x + 1) e^{x}

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

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