What is the solution for y^{''}-2y^'+y=(3x+1)e^x ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{2} - 13 d + 12) \cdot y = x

(d - 1)^{2} \cdot y = e^{x} \cdot sec^{2} (x) tan (x)

y^{^{''}} + y^{^{'}} - 2 y = 20 sin (2 x)

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

(d^{3} + 8) y = x^{4} + x + 1

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