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$(x-1)y''-xy'+y=0$

$y_1(x)=e^x$ is a solution of this differential equation, but how can I find a second linearly independent solution?

Here is what I've done so far..

$y_2(x)=ue^x$

$y_2'=u'e^x+ue^x$

$y_2''=u''e^x+u'e^x+u'e^x+ue^x$

Substituting back into the ODE, I get

$(x-1)(u''e^x+u'e^x+u'e^x+ue^x)-x(u'e^x+ue^x)+ue^x=0$

If I simplify this, I get

$x(u''+u'+2u)-u''-2u'=0$ ... Do you get the same?

Now what do I do, if this is correct?

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1 Answer

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$$ (x-1)(u''e^x+u'e^x+u'e^x+ue^x)-x(u'e^x+ue^x)+ue^x \\ = x(u''e^x+u'e^x) - (u''e^x+2u'e^x) + xu'e^x + xue^x- x(u'e^x+ue^x) - ue^x+ ue^x \\ = x(u''e^x+u'e^x) - (u''e^x+2u'e^x) $$

So solve

$$ (x-1)u'' + (x-2)u' = 0 $$

i.e. solve

$$ (x-1)w' + (x-2)w = 0 $$

for $w$, then $u$ is an antiderivative of $w$ and $y_2 = ue^x$.

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