What is the solution for y^{''}-2y^'+y=3x^2+6+4ex ?

The solution for y^{''}-2y^'+y=3x^2+6+4ex is y=c_{1}e^x+c_{2}xe^x+3x^2+(4e+12)x+8e+24

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y^{''}-2y^'+y=3x^2+6+4ex

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{2} + 6 + 4 e x

y = c_{1} e^{x} + c_{2} x e^{x} + 3 x^{2} + (4 e + 12) x + 8 e + 24

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 3 x^{2} + 6 + 4 e x

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 3 x^{2} + (4 e + 12) x + 8 e + 24

y = c_{1} e^{x} + c_{2} x e^{x} + 3 x^{2} + (4 e + 12) x + 8 e + 24

x^{^{''}} + x = 1, x (0) = 2, x^{^{'}} (0) = 1

y^{^{''}} + 9 y = 12 cos (3 x)

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

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

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

What is the solution for y^{''}-2y^'+y=3x^2+6+4ex ?
The solution for y^{''}-2y^'+y=3x^2+6+4ex is y=c_{1}e^x+c_{2}xe^x+3x^2+(4e+12)x+8e+24