What is the solution for linear y^{''}-2y^'+y=x^3+4x ?

The solution for linear y^{''}-2y^'+y=x^3+4x is y=c_{1}e^x+c_{2}xe^x+x^3+6x^2+22x+32

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linear y^{''}-2y^'+y=x^3+4x

linear

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 4 x

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 22 x + 32

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 22 x + 32

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 22 x + 32

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

2 x^{^{''}} + 50 x = cos (3 t)

y^{^{''}} + y = H (t) - H (t - \frac{π}{2})

y^{^{''}} + 4 y^{^{'}} + 5 y = 25 x^{2} + 13 sin (2 x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 4 e^{- 3 x} + 10 sin (x)

What is the solution for linear y^{''}-2y^'+y=x^3+4x ?
The solution for linear y^{''}-2y^'+y=x^3+4x is y=c_{1}e^x+c_{2}xe^x+x^3+6x^2+22x+32