What is the solution for (d^2+d)y=x^2+2x-3 ?

The solution for (d^2+d)y=x^2+2x-3 is y=c_{1}+c_{2}e^{-x}+(x^3)/3-3x

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(d^2+d)y=x^2+2x-3

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

y = c_{1} + c_{2} e^{- x} + \frac{x ^{3}}{3} - 3 x

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{x ^{3}}{3} - 3 x

y = c_{1} + c_{2} e^{- x} + \frac{x ^{3}}{3} - 3 x

2 y^{^{''}} + 12 y = 6 tan (3 x)

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{(- x)}

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

y^{^{''}} - 9 y = (x^{2} - 2) sin (3 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 11 e^{3} x, y^{1} (0) = e x

What is the solution for (d^2+d)y=x^2+2x-3 ?
The solution for (d^2+d)y=x^2+2x-3 is y=c_{1}+c_{2}e^{-x}+(x^3)/3-3x