Question Number 103037 by bramlex last updated on 12/Jul/20

4x^2 y′′ +y = 0 , x > 0

Answered by maths mind last updated on 12/Jul/20

let y=x^t ⇒4x^2 (t(t−1))x^(t−2) +x^t =0 ⇔(4t^2 −4t+1)x^t =0 ⇔(2t−1)^2 x^t =0⇒t=(1/2) let y=(√x) z ⇒y′=z′(√x)+(z/(2(√x))),y′′=((z′)/(√x))+z′′(√x)−(z/(4x(√x))) ⇔4x(√x)z′+4x^2 (√x)z′′−z(√x)+z(√x)=0 ⇒z′+xz′′=0 ⇒((z′′)/(z′))=−(1/x)⇒ln(z′)=−ln(x)+c ⇒z′=(k/x)⇒z=kln(x)+c ⇒y=(kln(x)+c)(√x)