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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + e^{x} x Ei (- x) + ln (x) - e^{x} Ei (- x) + 1

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + e^{x} x Ei (- x) + ln (x) - e^{x} Ei (- x) + 1

y = c_{1} e^{x} + c_{2} x e^{x} + e^{x} x Ei (- x) + ln (x) - e^{x} Ei (- x) + 1

y^{^{''}} = 2 y + 18 tan^{3} (x)

y^{^{''}} + 2 y^{^{'}} - 2 y = e^{- 2 x} (3 x^{2} + 5 x + 2)

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

5 y^{^{''}} + 25 y^{^{'}} + \frac{1}{1 0 ^{- 2}} y = 10 e^{- \frac{x}{4}}

y^{^{''}} + 2 y^{^{'}} - 35 y = 13 sin (x) - e^{3 x} + 1

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