What is the solution for y^{''}-2y^'+y=4x^2-6 ?

The solution for y^{''}-2y^'+y=4x^2-6 is y=c_{1}e^x+c_{2}xe^x+4x^2+16x+18

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y^{''}-2y^'+y=4x^2-6

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 6

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x + 18

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 4 x^{2} - 6

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x + 18

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x + 18

y^{^{''''}} - 2 y^{^{''}} + y = 5 sin (2 x)

y^{^{''}} + 3 y = e^{t}

y^{^{''}} - 9 y^{^{'}} = 9 x e^{- 3 x}

6 y^{^{''}} - 15 y^{^{'}} = 3 x^{2}

y^{^{''}} + 14 y^{^{'}} + 49 y = t - 2 e - 7 t

What is the solution for y^{''}-2y^'+y=4x^2-6 ?
The solution for y^{''}-2y^'+y=4x^2-6 is y=c_{1}e^x+c_{2}xe^x+4x^2+16x+18