I'm asked to
find a power series of the form $\sum a_n x^n$, recognize the resulting series as the expansion of a familiar function, then verify my conclusion by solving the equation directly.
Given $xy'=y$. However, I'm getting $a_0\sum_{n=0}^\infty (n! x^n)^{-1}$, which doesn't appear to match any functions that I know right now at least. (I was also expecting something involving $e$ just from looking at it.)
Here's what I got so far:
$$let \ \ y = \sum_{n=0}^\infty a_n x^n$$
$$y'=\sum_{n=1}^\infty na_nx^{n-1}$$
$$xy'=\sum_{n=1}^\infty na_nx^n$$
and since $xy'=y$:
$$\sum_{n=1}^\infty na_nx^n=\sum_{n=0}^\infty a_n x^n$$
$$\sum_{n=1}^\infty na_nx^n=\sum_{n=1}^\infty a_{n-1} x^{n-1}$$
$$na_nx^n=a_{n-1} x^{n-1}$$
$$a_nnx=a_{n-1}$$
$$a_n=\frac{a_{n-1}}{nx}$$
taking $a_0=a_0$
$$a_1=\frac{a_0}{x}, \qquad a_2=\frac{a_0}{2x^2} \qquad a_3=\frac{a_0}{6x^3}$$
and generally,
$$\sum_{n=0}^\infty \frac{a_0}{n!x^n}$$
Where did I go wrong?
The following line is correct $$\sum_{n=1}^\infty na_nx^n=\sum_{n=0}^\infty a_n x^n$$
$$\sum_{n=1}^\infty na_nx^n=a_0+\sum_{n=1}^\infty a_n x^n$$
$$-a_0+\sum_{n=1}^\infty a_nx^n(n-1)=0$$ $$\implies a_0=0$$ And $$\implies a_n(n-1)x^n=0 \implies n=1 \implies a_1 \ne 0 $$ $$ \implies n \ne 1 \implies a_n=0$$ The coefficent are zero except for $n=1$
Finally $$y=a_1x$$
Note that $$a_0\sum_{n=0}^\infty (n! x^n)^{-1}=a_0e^{1/x}$$ Since we have that $$\sum_{n=0}^\infty \frac {x^n}{n!}=e^x$$