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

The solution for y^{''}-2y^'+y=-4e^x+3x^2 is y=c_{1}e^x+c_{2}xe^x-2x^2e^x+3x^2+12x+18

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - 2 x^{2} e^{x} + 3 x^{2} + 12 x + 18

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - 2 x^{2} e^{x} + 3 x^{2} + 12 x + 18

y = c_{1} e^{x} + c_{2} x e^{x} - 2 x^{2} e^{x} + 3 x^{2} + 12 x + 18

y^{^{''}} + 3 y^{^{'}} + 4 y = 8 x + 2

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

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)

What is the solution for y^{''}-2y^'+y=-4e^x+3x^2 ?
The solution for y^{''}-2y^'+y=-4e^x+3x^2 is y=c_{1}e^x+c_{2}xe^x-2x^2e^x+3x^2+12x+18