Given that y_1 = e^{x}
y1 =e^x
is a solution of xy’’+(1-2x)y’+(x-1)y=0
xy’’+(1−2x)y’+(x−1)y=0, find its second solution y_2
y2
(0,∞), which is linearly independent from y1
y1=exy=y1∫u(x)dx=ex∫u(x)dxy′=ex∫u(x)dx+exu(x)y′′=ex∫u(x)dx+2exu(x)+exu′(x)x(ex∫u(x)dx+2exu(x)+exu′(x))+(1−2x)(ex∫u(x)dx+exu(x))+(x−1)ex∫u(x)dx=0xu′(x)+u(x)=0duu=−dxx⇒u=Cx⇒y2=ex∫1xdx=exlnxy_1=e^x\\y=y_1\int{u\left( x \right) dx}=e^x\int{u\left( x \right) dx}\\y'=e^x\int{u\left( x \right) dx}+e^xu\left( x \right) \\y''=e^x\int{u\left( x \right) dx}+2e^xu\left( x \right) +e^xu'\left( x \right) \\x\left( e^x\int{u\left( x \right) dx}+2e^xu\left( x \right) +e^xu'\left( x \right) \right) +\left( 1-2x \right) \left( e^x\int{u\left( x \right) dx}+e^xu\left( x \right) \right) +\left( x-1 \right) e^x\int{u\left( x \right) dx}=0\\xu'\left( x \right) +u\left( x \right) =0\\\frac{du}{u}=-\frac{dx}{x}\Rightarrow u=\frac{C}{x}\Rightarrow y_2=e^x\int{\frac{1}{x}dx}=e^x\ln x\\y1=exy=y1∫u(x)dx=ex∫u(x)dxy′=ex∫u(x)dx+exu(x)y′′=ex∫u(x)dx+2exu(x)+exu′(x)x(ex∫u(x)dx+2exu(x)+exu′(x))+(1−2x)(ex∫u(x)dx+exu(x))+(x−1)ex∫u(x)dx=0xu′(x)+u(x)=0udu=−xdx⇒u=xC⇒y2=ex∫x1dx=exlnx
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