I'm trying to solve the problem

$x*y' = 3*y + 3$

From the book Advanced Engineering Mathematics (By Erwin Kreyszig) - Chapter 5.2, exercise 1, using power series. I'm having trouble in this particular problem. This is what I've tried so far:

$x*y' - 3*y = 3$

Assume solution

$y = \sum_{m=0}^\inf a_m*x^m$


$y' = \sum_{m=1}^\inf m*a_m*x^{m-1}$

Which leads to

$x*(\sum_{m=1}^\inf m*a_m*x^{m-1}) - 3*(\sum_{m=0}^\inf a_m*x^m) = 3$

$\sum_{m=1}^\inf m*a_m*x^{m} + \sum_{m=0}^\inf -3*a_m*x^m = 3$

Assuming any $a_0$, I know that $m * a_0 * x^0 = 0$, for $m = 0$, then it is possible to rewrite the first sum:

$\sum_{m=0}^\inf m*a_m*x^{m} + \sum_{m=0}^\inf -3*a_m*x^m = 3$

Merging the two series:

$\sum_{m=0}^\inf (m - 3)*a_m*x^{m} = 3$

For $x=0$, we know that

$\sum_{m=0}^\inf (m - 3)*a_m*x^{m} = (0-3)*a_0*0^0 = -3*a_0$

$-3*a_0 = 3$

$a_0 = -1$

And that's it. I'm stuck because the rest of the series doesn't seem to relate $a_n$ with $a_0$, and since it is not homogeneous, it is not possible that $a_0 = a_1 = a_2 = ... = 0$. Also, this series doesn't seem to be convergent anyway, so even if I manage to find all $a_n$, depending on their values, I may not get a solution...

Any help at this point would be nice. Thanks in advance.


1 Answer 1


Everything's fine so far, since what you have is actually $$ \sum_{m=0}^\infty (m-3) a_m x^m = 3 + 0 x + 0 x^2 + 0 x^3 + \dots $$ So you can compare coefficients term by term to get $-3a_0=3$, $-2a_1=0$, $-a_2=0$, $0a_3=0$, etc.

(As a check, you can also compare with the solution that you get using an integrating factor: $(x^{-3} y)' = 3 x^{-4} \iff \ldots$)

  • $\begingroup$ How did you get $a_n = 0$, $n\ge1$ ? I thought all other terms were $0$ because we choose $x=0$. $\endgroup$ Commented Oct 29, 2017 at 19:17
  • $\begingroup$ Because the right-hand side in the equation $xy'-3y=3$ is the constant function $3$, whose Maclaurin series is $3+0x+0x^2+\dots$. If you just look at the value at the point $x=0$, you can't say anything. The crucial thing is that if two functions are the same (for all $x$), then their power series must of course be the same too. $\endgroup$ Commented Oct 29, 2017 at 19:21
  • $\begingroup$ I was struggling to notice that the series have the term $(3-3)*a_3*x^3$, so it seems that $a_3$ can be any value, and the final series is $y=\sum_{m=0}^{\inf}a_m*x^m=a_0*x^0 + a_3*x^3$ (Since all other terms are 0), and, since $a_0=-1$, the solution is $y=-1 + a_3*x^3$. Thanks for the help! :) $\endgroup$ Commented Oct 29, 2017 at 20:11
  • $\begingroup$ That's correct. You're welcome! $\endgroup$ Commented Oct 29, 2017 at 20:24

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