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

The solution for y^{''}-2y^'+y=x(e^x)+4 is y=c_{1}e^x+c_{2}xe^x+(e^xx^3)/6+4

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6} + 4

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6} + 4

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6} + 4

D (x) = 120 x - 3580

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

y^{^{''}} - y^{^{'}} + y = 3 sin (t) - 9 e^{2 t}

y^{^{''}} - 2 y + 37 y = e^{x} cos (6 x)

d, d (t) = 60 - 4.9 t \cdot 2

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