What is the solution for y^{''}-2y^'+y=xe^x+4,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-2y^'+y=xe^x+4,y(0)=1,y^'(0)=1 is y=-3e^x+4e^xx+(e^xx^3)/6+4

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y^{''}-2y^'+y=xe^x+4,y(0)=1,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + y = x e^{x} + 4, y (0) = 1, y^{^{'}} (0) = 1

y = - 3 e^{x} + 4 e^{x} x + \frac{e ^{x} x ^{3}}{6} + 4

Solution steps

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

Solve linear ODE:

y = - 3 e^{x} + 4 e^{x} x + \frac{e ^{x} x ^{3}}{6} + 4

y = - 3 e^{x} + 4 e^{x} x + \frac{e ^{x} x ^{3}}{6} + 4

y^{^{''}} - 6 y = u^{2} (t), y (0) = 0

y^{^{''}} + 16 y = e^{5 t}

y^{^{''}} + 256 y = 16 cos (3 x)

D (x) = - 6 (2940) + 4200

y^{^{''''}} - 16 y = 2 e^{- 2 x} + 3 e^{3 x} + cos (2 x) - 1

What is the solution for y^{''}-2y^'+y=xe^x+4,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-2y^'+y=xe^x+4,y(0)=1,y^'(0)=1 is y=-3e^x+4e^xx+(e^xx^3)/6+4