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## MathSofiya 3 years ago Use power series to solve the differential equation $(x-1)y''+y'=0$ $y=\sum_{n=0}^\infty a_nx^n$ $y'=\sum_{n=1}^\infty na_nx^{n-1}$ $y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}$ $(x-1)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0$

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1. MathSofiya

allow me to complete this problem as far as I can, give me like 5 mins. Thanks!

2. MathSofiya

$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0$

3. MathSofiya

$\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n-\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^\infty (n+1)a_{n+1}x^n=0$

4. MathSofiya

ok so when we strip out a term on $\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n$ for the sum to start at n=0 would I have... $a_0+ \sum_{n=0}^\infty n(n+1)a_{n+1}x^n$

5. MathSofiya

no that would be $a_0+ \sum_{n=1}^\infty n(n+1)a_{n+1}x^n$

6. mahmit2012

cof of x^n (left)=cof of x^n (right) don't use sigmas.

7. mahmit2012

already I have showed you ! Take a look to my solution.

8. mahmit2012

y=an y'=(n+1)an+1 xy'=nan y''=(n+2)(n+1)an+2 xy''=(n+1)nan+1 x2y=n(n-1)an and so on.

9. mahmit2012

(n+1)nan+1-(n+2)(n+1)an+2+(n+1)an+1=0 an+2=(n+1)/(n+2) an+1 an+2=a1/(n+2) one answer is just a0(check it out) y=a0+a1sigma(n=1 to inf)(x^n/n)

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