What is the solution for y^{''}-y^'=0+2t^2+t-1 ?

The solution for y^{''}-y^'=0+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^'=0+2t^2+t-1

y^{^{''}} - y^{^{'}} = 0 + 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) = 0 + 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^{^{''}} - y^{^{'}} + y = 25 sin (2 x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{( 1 - x ) ^{2}}

x^{^{''}} = - a^{2} x + b

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 t} cos (5 t)

a^{2} y^{^{''}} - (a^{2} + 1) y^{^{'}} + y = x

What is the solution for y^{''}-y^'=0+2t^2+t-1 ?
The solution for y^{''}-y^'=0+2t^2+t-1 is y=c_{1}e^t+c_{2}-(2t^3)/3-(5t^2)/2-4t