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

midtlinjeg

midtlinjeg

Answered question

2021-02-12

Use Taylor series expansion to find the third degree power series solution of the initial value problem
(x1)y+2y4y=0,y(0)=2,y(0)=6

Answer & Explanation

Tasneem Almond

Tasneem Almond

Skilled2021-02-13Added 91 answers

Given that:
(x1)y+2y4y=0,y(0)=2,y(0)=6
Let y=n=0anxn be a power series expansion of the required solution.
y=n=0anxn
y=n=0an(n)xn1=a1+2a2x+3a3x2+...
y=n=0ann(n1)xn2=2a2+6a3x+12a4x2+20a5x3+...
Now, find the Taylor series expansions of the given series:
(x1)y+2y4y=0,y(0)=2,y(0)=6
y(0)=2a0=2
y(0)=6a1=6
(x1)y+2y4y=0

(x1)(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+...)=0
Compare the coefficient of x0,x1 and power:
(2a2+128)+(2a26a3+4a224)x+...
2a2+128=02a2=12+8a2=2
Coefficient of x:
2a26a3+4a224=0
2a26a3+4a2=24

Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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