"similitude method" for the case: $\nu''-2\nu'+\nu= \frac {e^x}{x}$ As title says... I have already tried things like $\nu(x)=\log(x^{\lambda})e^x$ or $\nu(x)=\dfrac{e^x \lambda}{x}$ But with no success, stinks of logarithms but I don't know...

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

$$y''-y'-(y'-y)= \frac {e^x} x$$ Substitute $h=y'-y$ $$h'-h=\frac {e^x} x$$

Which is a first ODe easy to solve

$$ {e^{-x}h'-e^{-x}h}=\frac {1} x$$

$$ {h}{e^{-x}}=\int \frac {dx} x$$

$$ h=y'-y= {e^{x}}(\ln(x)+K)$$

$$ e^{-x}y'-ye^{-x}= \ln(x)+K$$

$$ (ye^{-x})'= \ln(x)+K$$

$$ y=e^{x} \int(\ln(x)+K )dx$$

$$ y=e^{x}(x\ln(x)+K_1x+K_2)$$