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How to solve recurrence relation $y_n= y_{n-1}+ (n-1).y_{n-2}$ where $n$ is a variable ? $y_n$ is a $n$th term, $y_{n-1}$ is $(n-1)$th term and $y_{n-2}$ is $(n-2)$th term.

soroosh.strife
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user208872
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1 Answers1

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Hint

Let $y_{0}=a,y_{1}=b$,and let $$f(x)=\sum_{n=0}^{\infty}y_{n}x^n$$ then we have $$y_{n}x^n=xy_{n-1}x^{n-1}+x^2(n-2)y_{n-2}x^{n-2}+y_{n-2}x^n$$ $$\Longrightarrow \sum_{n=2}^{\infty}y_{n}x^n=x\sum_{n=2}^{\infty}y_{n-1}x^{n-1}+x^2\sum_{n=2}^{\infty}(n-2)y_{n-2}x^{n-2}+x^2\sum_{n=2}^{\infty}y_{n-2}x^{n-2}$$ $$\Longrightarrow f(x)-a-bx=x[f(x)-a]+x^3f'(x)+x^2f(x)$$ so $$f'(x)+\dfrac{x^2+x-1}{x^3}f(x)=\dfrac{(a-b)x-a}{x^3}$$ so $$f(x)=e^{-\int\frac{x^2+x-1}{x^3}dx}\left(\int\dfrac{(a-b)x-a}{x^3}e^{\int\frac{x^2+x-1}{x^3}dx}dx+C\right)$$ then I think you can do it,even it's ugly

math110
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  • Why F(x) is solution for given recurrence relation. please explain. – user208872 Jan 22 '15 at 06:17
  • @user208872, f(x) is not a final solution. Final solution you get after writing the coefficients of consecutive powers of x. – banan3'14 Jun 15 '20 at 13:53
  • Is it doable from the point when we solved the differential equation? We still have to solve for $C$, but how can we construct the initial value problem? Obviously $x \ne 0$, but if we take $x=1/2$ for example, then we don't know $f(1/2)$. It seems impossible to me. – banan3'14 Jun 15 '20 at 14:09