Use Taylor series expansion to ﬁnd the third degree power series solution of the initial value problem (x−1)y′′+2y′−4y=0,y(0)=2,y′(0)=6

Question

Use Taylor series expansion to ﬁnd the third degree power series solution of the initial value problem  (x−1)y′′+2y′−4y=0,y(0)=2,y′(0)=6

Tasneem Almond · Accepted Answer

Given that:(x−1)y′′+2y′−4y=0,y(0)=2,y′(0)=6Let y=∑n=0∞anxn be a power series expansion of the required solution.y=∑n=0∞anxny′=∑n=0∞an(n)xn−1=a1+2a2x+3a3x2+...y″=∑n=0∞ann(n−1)xn−2=2a2+6a3x+12a4x2+20a5x3+...Now, find the Taylor series expansions of the given series:(x−1)y′′+2y′−4y=0,y(0)=2,y′(0)=6y(0)=2⇒a0=2y′(0)=6⇒a1=6(x−1)y′′+2y′−4y=0⇒(x−1)(2a2+6a3x+12a4x2+20a5x3+...)+2(a1+2a2x+3a3x2+...)−4(a0+a1x+a2x2+a3x3+...)=0⇒(2a2x+6a3x2+12a4x3+...)−(2a2+6a3x+12a4x2+20a5x3+...)+(2a1+4a2x+6a3x2+...)−(4a0+4a1x+4a2x2+4a3x3+...)=0⇒(2a2x+6a3x2+12a4x3+...)−(2a2+6a3x+12a4x2+20a5x3+...)+(12+4a2x+6a3x2+...)−(8+24x+4a2x2+4a3x3+...)=0Compare the coefficient of x0,x1 and power:(−2a2+12−8)+(2a2−6a3+4a2−24)x+...−2a2+12−8=0⇒−2a2=−12+8⇒a2=2Coefficient of x:2a2−6a3+4a2−24=02a2−6a3+4a2=24

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Use Taylor series expansion to ﬁnd the third degree power series solution of the initial value problem (x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6

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