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

The solution for y^{''}-2y^'+y=(e^x)/x ,y(1)=0,y^'(1)=1 is y=(-1/e+1)e^x+(1-e)e^{x-1}x+e^xxln(x)

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

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

y = (- \frac{1}{e} + 1) e^{x} + (1 - e) e^{x - 1} x + e^{x} x ln (x)

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{e} + 1) e^{x} + (1 - e) e^{x - 1} x + e^{x} x ln (x)

y = (- \frac{1}{e} + 1) e^{x} + (1 - e) e^{x - 1} x + e^{x} x ln (x)

y^{^{''}} + y^{^{'}} + 4 y = e^{- 1.3 t}

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

y^{^{''}} - 2 y^{^{'}} + y = 30 e^{t} t^{- 7}

y^{^{''}} + 4 y = 12 t

y^{^{''}} - y^{^{'}} + 2 y = 2 e^{2 t} + 5 t - 1

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