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

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
Series
asked 2021-02-12
Use Taylor series expansion to find the third degree power series solution of the initial value problem
\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)

Answers (1)

2021-02-13
Given that:
\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)
Let \(y=\sum_{n=0}^\infty a_nx^n\) be a power series expansion of the required solution.
\(y=\sum_{n=0}^\infty a_nx^n\)
\(y'=\sum_{n=0}^\infty a_n(n)x^{n-1}=a_1+2a_2x+3a_3x^2+...\)
\(y''=\sum_{n=0}^\infty a_nn(n-1)x^{n-2}=2a_2+6a_3x+12a_4x^2+20a_5x^3+...\)
Now, find the Taylor series expansions of the given series:
\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)
\(y(0)=2\Rightarrow a_0=2\)
\(y'(0)=6\Rightarrow a_1=6\)
\((x−1)y′′+2y′−4y =0\)
\(\Rightarrow(x-1)(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+2(a_1+2a_2x+3a_3x^2+...)-4(a_0+a_1x+a_2x^2+a_3x^3+...)=0\)
\(\Rightarrow(2a_2x+6a_3x^2+12a_4x^3+...)-(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+(2a_1+4a_2x+6a_3x^2+...)-(4a_0+4a_1x+4a_2x^2+4a_3x^3+...)=0\)
\(\Rightarrow(2a_2x+6a_3x^2+12a_4x^3+...)-(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+(12+4a_2x+6a_3x^2+...)-(8+24x+4a_2x^2+4a_3x^3+...)=0\)
Compare the coefficient of \(x^0,x^1\) and power:
\((-2a_2+12-8)+(2a_2-6a_3+4a_2-24)x+...\)
\(-2a_2+12-8=0\Rightarrow-2a_2=-12+8\Rightarrow a_2=2\)
Coefficient of x:
\(2a_2-6a_3+4a_2-24=0\)
\(2a_2-6a_3+4a_2=24\)
\(\Rightarrow2(2)-6a_3+4(2)=24\)
\(\Rightarrow-6a_3=-4(2)+24-2(2)\Rightarrow a_3=2\)
The third degree power series solution is:
\(y(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+...\)
\(y(x)=a_0+a_1x+a_2x^2+a_3x^3+...\)
\(y(x)=2+6x+2x^2+2x^3+...\)
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