Question #320988


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


1
Expert's answer
2022-03-31T06:41:11-0400

y1=exy=y1u(x)dx=exu(x)dxy=exu(x)dx+exu(x)y=exu(x)dx+2exu(x)+exu(x)x(exu(x)dx+2exu(x)+exu(x))+(12x)(exu(x)dx+exu(x))+(x1)exu(x)dx=0xu(x)+u(x)=0duu=dxxu=Cxy2=ex1xdx=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\\


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