Answer to Question #347065 in Differential Equations for Daniel

Differential Equations

Question #347065

Show that 𝑦 = 2𝑥 ln 𝑥 is a solution to the differential equation "x^2y''-xy'+y=0"

Expert's answer

"x^2y''-xy'+y=0"

"y=2x\\ln x"

"y'=2\\ln x+2"

"y''=2\/x"

"x^2y''-xy'+y=\\\\\n=x^2(2\/x)-2x\\ln x-2x+2x\\ln x=0"

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