Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

By finding solutions as power series in $x$ solve $$4xy''+2(1-x)y'-y=0 .$$ What I did is the following. First I let the solution $y$ be equal to $$y =\sum_{i=0}^{\infty} b_ix^i =b_0 +b_1x+b_2x^2+\ldots$$ for undetermined $b_i$. Then I found the expression for $y'$ and $y''$, $$y' =\sum_{i=0}^{\infty} ib_ix^{i-1} =b_1 + 2b_2x+3b_3x^2+\ldots.$$ and $$y'' =\sum_{i=0}^{\infty} i(i-1)b_ix^{i-2} =2b_2+6b_3x+12b_4x^2\ldots.$$ Now I put these in the original DE to get $$4\sum i(i-1)b_ix^{i-1}+2\sum ib_i(x^{i-1}-x^i) - \sum b_ix^i =0 $$ where all sums range from $0$ to infinity. Finally this becomes $$\sum \left\{ (4i(i-i)b_i+2ib_i )x^{i-1}+(-2ib_i-b_i)x^i \right\}=0.$$ At this point I am fairly certain I have already made a mistake somewhere, probably in working out the power series of $y'$ or $y''$. Who can help point it out to me, I am pretty sure in the last sum there should be terms like $b_{i+1}$ or $b_{i+2}$. Thanks for any help or tips!

EDIT I have gotten further by realizing that $$y' =\sum_{i=0}^{\infty} ib_ix^{i-1} =\sum_{i=1}^{\infty} ib_ix^{i-1}=\sum_{i=0}^{\infty} (i+1)b_{i+1}x^{i}$$ and $$y'' =\sum_{i=0}^{\infty} (i+2)(i+1)b_{i+2}x^{i}.$$ Putting these in the original DE I get $$\sum \left\{ [4(i+2)(i+1)b_{i+2}-2(i+1)b_{i+1}]x^{i+1} + [2(i+1)b_{i+1}-b_i]x^i \right\}=0.$$ This must be true for all $x$ and thus we have $$4(i+2)(i+1)b_{i+2}=2(i+1)b_{i+1}$$ and $$2(i+1)b_{i+1} = b_i.$$ After simplyfying these two conditions are seen to be identical. Now I've set $b_0=1$ to obtain the solution $$ y = 1 + \frac{x}{2}+ \frac{x^2}{8} +\frac{x^3}{48}+\ldots + \frac{x^i}{2^i(i!)}+\ldots.$$ Now I've arrived at the ackward position where in working out the question here I have actually managed to solve it. My last question is then, does anyone recognize this power series? Thanks!

share|improve this question
1  
Yes you solved it. Next you might want to post your solution as an answer (this is even recommanded, and is surely better than to accept a flawed answer). About your last power series, surely you would recognize $\sum\limits_i\frac{x^i}{i!}$? Compare to your case... –  Did Jan 5 at 13:41
    
Ah thanks a lot for the tip a the end. I do indeed recognize that series :) so the answer is $e^{\frac{x}{2}}$ or if I had chosen an arbitrary $b_0 =C$ I would have $Ce^{\frac{x}{2}}$, however, shouldn't I have found two independent solutions? –  Slugger Jan 5 at 14:12
    
Indeed, but the other solution is not a series in x, hence one cannot "see" it with this approach. –  Did Jan 5 at 14:17

2 Answers 2

up vote 4 down vote accepted

You have made your mistake in the power series. In particular, you need to end up with a recurrence relation and solve that. $$y'=\sum_{i=0}^\infty{ib_ix^{i-1}}=0+b_1+2b_2x+3b_3x^2+...=\sum_{i=1}^\infty{ib_ix^{i-1}}$$ Now you need to get your lower bound so that it starts at $0$. Rewriting the sum using $i=0$, we get that $$\sum_{i=1}^\infty{ib_ix^{i-1}}=\sum_{i=0}^\infty{(i+1)b_{i+1}x^i}$$ Similarly, $$y''=\sum_{i=0}^\infty{i(i-1)b_ix^{i-2}}=0(-1)x^{-2}+1(0)b_1x^{-1}+2(1)b_2+3(2)b_3x+...=\sum_{i=2}^\infty{ib_ix^{i-2}}$$ Now rewrite that also with an index of 0. $$\sum_{i=2}^\infty{i(i-1)b_ix^{i-2}}=\sum_{i=0}^\infty{(i+2)(i+1)b_{i+2}x^i}$$ Since all the indices are now $0$, you can rewrite the equation as
$$4x\sum_{i=0}^\infty{(i+2)(i+1)b_{i+2}x^i}+2(1-x)\sum_{i=0}^\infty{(i+1)b_{i+1}x^i}-\sum_{i=o}^\infty{b_ix^i}=0$$ You now have one more issue to resolve. You have to include the factors of $x$ in both the $y''$ and $y'$ sums and this gives you two higher powers of $x$. You'll again have to rewrite the sums so that each sum contains sums of $x^i$, not $x^{i+1}$.
$$\sum_{i=0}^\infty{4i(i+1)b_{i+1}x^i}+\sum_{i=0}^\infty{2(i+1)b_{i+1}x^i}-\sum_{i=o}^\infty{2ib_ix^i}-\sum_{i=0}^\infty{b_ix^i}$$ $$=\sum_{i=0}^\infty{[2(2i+1)(i+1)b_{i+1}-(2i+1)b_i}]x^i$$ So, we then see that $2(2i+1)(i+1)b_{i+1}=(2i+1)b_i$ and thus $$2(i+1)b_{i+1}=b_i\Rightarrow b_{i+1}=\frac{b_i}{2(i+1)}$$ Setting $b_0=1$, and replacing the $b_i$'s in the series expansion of $y$, we get $$y=\sum_{i=0}^\infty{\frac{x^i}{2^ii!}}=\sum_{i=0}^\infty{\frac{(\frac{x}{2})^i}{i!}}=e^{\frac{x}{2}}$$

share|improve this answer
    
"you can rewrite the equation as one sum which has terms $b_i$, $b_{i+1}$, and $b_{i+2}$" I doubt that. Only consecutive coefficients are involved, say $b_i$ and $b_{i+1}$. –  Did Jan 5 at 13:33
    
Thanks a lot for your answer, I actually recognized my mistake and worked out my answer. Still thanks a lot for your answer! –  Slugger Jan 5 at 13:40
    
Answer still wrong (if that matters). –  Did Jan 5 at 13:42
    
I just put the x into the $y''$ and reduced the $b_{i+2}$ to $b_{i+1}$ so i see where you are leading to, @Did. But is my answer wrong or just not finished? –  Eleven-Eleven Jan 5 at 13:46
1  
Ambiguous only in the revised version, since, inserting the prefactors x and 1-x in the series, the coefficient of x^i involves b_i and b_{i+1} only. –  Did Jan 5 at 14:02

$$\sum_i(4i(i-1)b_i+2ib_i )x^{i-1}+(-2ib_i-b_i)x^i=\sum_i(2(i+1)b_{i+1}-b_i)(2i+1)x^i$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.