I want to solve this differential equation with the power-series method: $$ x^{2}\cdot y''(x) +(1+x^{2}) y'(x) + y(x) =0$$ where $$y(0)=1$$

They want the solution given in elementary functions. I managed to get the recursive formula for $$j\ge 2$$$$a_{j+1} = \frac{-((j(j-1)+1)a_{j} + (j-1)a_{j-1})}{j+1} $$ And also got $$a_{0}=y(0)=1 $$,and then$$ a_{1}=-1, a_{2}=1/2 $$

But Im not sure how to answer in an elementary function, I tried to get some more values and its just irrational values. Appreciate help.


I agree with you're recurrence relation $(i+1)a_{i+1}=-((i^2-i+1)a_i+(i-1)a_{i-1})$. & the first few values $a_0=1, a_1=-1, a_2=1/2, a_3=-1/6$ from this I guess the general formulae $a_i=(-1)^i/i!$. You can easily show (by induction) that this is indeed true. Now the solution is obviously $y=e^{-x}$ & this does indeed satisfy the original equation.

  • $\begingroup$ But the value I get for $$ a_{4}=-13/24 $$? This doesnt make sense? $\endgroup$ – fejz1234 Jan 29 '17 at 20:18
  • $\begingroup$ Or my bad it is equal to 1/24 actually, but how does one deduce that it is equal to e^(-x) ? $\endgroup$ – fejz1234 Jan 29 '17 at 20:21
  • $\begingroup$ Oh ... let me see what I get ? $4a_4=-(7a_3+2a_2)$ ... $a_4=1/24$ ? $\endgroup$ – Donald Splutterwit Jan 29 '17 at 20:22
  • $\begingroup$ I mean how do you go from $$ \frac{(-1)^{i}}{i!} = e^{-x} $$ , is taylor-expansion that reveals it? $\endgroup$ – fejz1234 Jan 29 '17 at 20:23
  • $\begingroup$ Thats right \begin{eqnarray*} \sum_{i=0}^{\infty} \frac{(-1)^i}{i!}x^i =e^{-x} \end{eqnarray*} $\endgroup$ – Donald Splutterwit Jan 29 '17 at 20:25

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