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

The solution for y^{''}-2y^'+y=4x^2-3 is y=c_{1}e^x+c_{2}xe^x+4x^2+16x+21

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x + 21

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x + 21

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x + 21

\frac{1}{2} y^{^{''}} + \frac{1}{2} y^{^{'}} + 6 y = 10 cos (3 x)

y^{^{''}} + \frac{750}{30} y = \frac{20}{30} cos (2) t

5 y^{^{''}} + 180 y = 150 sin (3 t)

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

y^{^{''}} + 25 y = cos (9 x)

What is the solution for y^{''}-2y^'+y=4x^2-3 ?
The solution for y^{''}-2y^'+y=4x^2-3 is y=c_{1}e^x+c_{2}xe^x+4x^2+16x+21