What is the solution for y^{''}-2y^'+y=12x^3e^x ?

The solution for y^{''}-2y^'+y=12x^3e^x is y=c_{1}e^x+c_{2}xe^x+(3e^xx^5)/5

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y^{''}-2y^'+y=12x^3e^x

y^{^{''}} - 2 y^{^{'}} + y = 12 x^{3} e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{3 e ^{x} x ^{5}}{5}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 12 x^{3} e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{3 e ^{x} x ^{5}}{5}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{3 e ^{x} x ^{5}}{5}

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{2 t}

y^{^{'''}} - 6 y^{^{''}} - 6 = cos (x)

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 3 + 3 x

y^{^{'''}} + 4 y^{^{'}} = e^{x} + sin (x)

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

What is the solution for y^{''}-2y^'+y=12x^3e^x ?
The solution for y^{''}-2y^'+y=12x^3e^x is y=c_{1}e^x+c_{2}xe^x+(3e^xx^5)/5