# Question

Find the first two non-zero terms of each of two power series solution about the origin of $$e^xy''+xy=0$$

# Attempt

Let $$y= \sum_{n=0}^\infty a_nx^n$$ Substituting $y$ in the given equation gives. $$e^xy''+xy = \sum_{n=0}^\infty \frac{x^n}{n!}\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n +\sum_{n=0}^\infty a_nx^{n+1}=0$$

Setting the coefficients to $0$ gives us $$a_2 = 0;~~~ a_3 = -\frac{1}{6}a_0;~~~ a_4 = \frac{1}{12}\left(a_0-a_1\right)$$

Substituting it back into $y$ gives us $$y = a_0 +a_1x-\frac{1}{6}a_0 x^3 +\frac{1}{12}\left(a_0-a_1\right) x^4$$

However the provided solution is $$y = 1 -\frac{1}{6} x^3 +\frac{1}{12} x^4+...$$

$$y = x +\frac{1}{12} x^4+...$$

Why are we supposed to assume $(a_0,a_1) = (1,0)$ for one solution and $(a_0,a_1) = (0,1)$?

Is it because the solution set is a vector space with the basis $(1,0), (0,1)$?

Moreover, would such assumption (that the solution set is a vector space and hence can be written as linear combination of $y_1$ and $y_2$) be feasible if there was a value of $n$ for which $a_n$ was non-linear in $a_0$ or $a_1$ or at the very least, was $ca_0a_1$? (Since the coefficient would no longer remain and hence would not be a vector space

In my opinion the wording is inappropriate. There are more that two series solutions to the equation. It would be better to ask for two linearly independent solutions. Then, because of linearity, you can take as initial conditions any pair $(a_0,a_1)$, $(b_0,b_1)$ that are linearly independent; $(1,0)$ and $(0,1)$ seem the simplest choice.