I am currently working on some introductory problems for series solutions for ODEs and am really struggling. The question is as follows:

$$ (7+x)y' = y $$ Calculate the first five terms in the series. So far I have done the following but am unsure how to continue.

Let $ y = \sum_{m=0}^\infty a_mx^n$ and $y' = \sum_{m=1}^\infty ma_mx^{n-1}$

I have substitute those back into the DE but get a little lost when I start manipulating the expression and changing the indices of the sums but so far I believe my best attempt is:

$$(7+x)2a_1 + \sum_{m=0}^\infty a_{m+1}(m+1) - a_{m-1} = 0$$

I am not convinced what I have done to get to this point is correct but I have tried to carry on and come up with expressions for the coefficients of the series but am just not understanding it. I have read around through similar questions and trying to further understand the topic but would appreciate any help coming up with just the first couple of terms in the series to get me started.

Thank you,


  • $\begingroup$ looks like $sinh(ln(x+7))+cosh(ln(x+7))$ would do it - or am i wrong? (you could calculate the coefficients easily then) $\endgroup$ – Max May 31 '16 at 12:23
  • $\begingroup$ You are mixing $m$ and $n$ in your second equation. $\endgroup$ – MrYouMath May 31 '16 at 12:31

You want all the series to be written with the "same power" $x^m$. Then $$ y'=\sum_{m=0}^\infty m\,a_m\,x^{m-1}=\sum_{m=0}^\infty (m+1)\,a_{m+1}\,x^{m}, $$ $$ x\,y'=\sum_{m=0}^\infty m\,a_m\,x^{m}. $$ Putting it all together you get $$ =\sum_{m=0}^\infty\bigl(7\,(m+1)\,a_{m+1}+m\,a_m-a_m\bigr)x^m=0. $$

  • $\begingroup$ Thank you I have carried on and think I have correctly evaluated the series as I have a0 + a0/7 + 0 + 0 + 0. Thank you very much $\endgroup$ – Michael May 31 '16 at 13:08

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