Answer to Question #320988 in Differential Equations for Aps

Question

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

Answer 1

$y_1=e^x\\y=y_1\int{u\left( x \right) dx}=e^x\int{u\left( x \right) dx}\\y'=e^x\int{u\left( x \right) dx}+e^xu\left( x \right) \\y''=e^x\int{u\left( x \right) dx}+2e^xu\left( x \right) +e^xu'\left( x \right) \\x\left( e^x\int{u\left( x \right) dx}+2e^xu\left( x \right) +e^xu'\left( x \right) \right) +\left( 1-2x \right) \left( e^x\int{u\left( x \right) dx}+e^xu\left( x \right) \right) +\left( x-1 \right) e^x\int{u\left( x \right) dx}=0\\xu'\left( x \right) +u\left( x \right) =0\\\frac{du}{u}=-\frac{dx}{x}\Rightarrow u=\frac{C}{x}\Rightarrow y_2=e^x\int{\frac{1}{x}dx}=e^x\ln x\\$