# Series solution to first order differential equation

I need to find a series solution to the following simple differential equation $$x^2y'=y$$ Assuming the solution to be of the form $y=\sum a_nx^n$ and equating the coefficients on both the sides, all the coefficients turn out to be zero which is definitely wrong.

Please guide me to the correct solution.

Any help is appreciated. Thanks!

• Why would you want a series solution for such a simple problem? Is that the required? – Hasan Saad Mar 13 '15 at 12:25
• If you do a series solution about x = 0, a standard one will fail, since $x=0$ is a singular point. – Paul Mar 13 '15 at 12:37
• @Paul: How do you know that $x=0$ is a singular point? – Pranav Arora Mar 13 '15 at 12:57
• @PranavArora, if you divide through by $x^2$, so that the coefficient of the highest derivative is 1, then the equation is not defined at $x=0$. – Paul Mar 13 '15 at 13:48
• differential equation jas integrable singularity at $x = 0.$ – abel Mar 13 '15 at 19:19

## 3 Answers

if you let $y = \sum_{ n = 0}^{\infty} a_nx^n$

$x^2y' = \sum_{ n = 0}^{\infty} a_nnx^{n+1} = \sum_{ n = 0}^{\infty} a_nx^n$

so equating coefficients on each side gives you $a_{0} = 0$ and a relation

$$a_{n+1} = na_n$$

which does indeed lead to all coefficients being zero

What if you supposed that $y = \sum_{ -\infty}^{\infty} a_nx^n$

then $a_{-0}$ could equal A say and then you would get $-1a_{-1} = A$

You should then be able to recover the answer that Autolatry found using separation of variables

• I did suppose the other series you mention and it did give the right answer but I do not see how this choice of series can be obvious to someone. Thanks! – Pranav Arora Mar 13 '15 at 13:10
• Because if x = 0 then y = 0 but y' is not defined at that point (it can take any value) and neither is any higher derivative. The power series expansion from Taylor's theorem requires these derivatives to exist around the point of expansion. Really the question is the other way round. Can the use of only positive powers of x be justified without considering negative powers? – Michael E Mar 13 '15 at 13:25
• I think the intuition for the choice though probably comes from Laurent series in complex analysis. There the co-efficient of x^-1 is of particular importance. – Michael E Mar 13 '15 at 13:28
• I am not sure if I understood what you wrote so this could really sound as a silly question to you. How do you know that y' is not defined at that point? – Pranav Arora Mar 13 '15 at 15:44

It could be useful to remark that $y(x)=0$ is the only solution of $x^2 y^\prime (x)=y(x)$ which is defined and analytic on the whole real line. In fact, the other solutions of your ODE are of the type: $$y(x)=C \exp \left( -\frac{1}{x}\right)$$ and none of them is both defined and analytic in $\mathbb{R}$.

Therefore, it is really obvious that the power series method yields only the zero solution.

Moreover, if you let: $$y(x) := u\left( \frac{1}{x}\right)$$ (for a suitable differentiable $u(t)$) you can reduce your ODE to a simpler one: in fact, you have: $$y^\prime (x) = -\frac{1}{x^2}\ \dot{u}\left( \frac{1}{x}\right)$$ (where the dot denotes derivation with respect to the $t$ variable) that is: $$\dot{u} \left( \frac{1}{x}\right) = -x^2\ y^\prime (x)\; .$$ Therefore $y(x)$ solves your ODE away from zero iff $u(t)$ solves: $$\dot{u}(t) = - u(t)\; .$$ Hence $u(t)=Ce^{-t}$ and $y(x)=Ce^{-1/x}$.

Well, \begin{eqnarray} \int \frac{dy}{y} &=& \int \frac{dx}{x^{2}}\\ \implies \ln y &=& -\frac{1}{x}+c \end{eqnarray} Hence, \begin{equation} y=Ae^{-\frac{1}{x}} \end{equation} Is a series solution necessary?

• $y=0$ is also a solution. – Paul Mar 13 '15 at 12:40
• @Paul Of course. – Autolatry Mar 13 '15 at 12:43
• I know how to solve a first order differential equation but finding the series solution is an exercise in my book. – Pranav Arora Mar 13 '15 at 12:57
• Series solutions are usually the method of choice for second order differential equations; I don't see the reason behind using that method here; could the ODE be $x^{2}y''=y$? – Autolatry Mar 13 '15 at 13:01
• Actually, if indeed the initial equation is a second order, then it is a form of Euler's equation written as $x^{2}y^{''}+0\cdot xy^{'}-y=0$. – Autolatry Mar 13 '15 at 13:08