How to solve this differential using only power series?

Question

I am asked to solve this differential equation using only power series. I'm really stuck. How do I even start this?

$$y' - y = 0$$

I think the answer is $e^{-x}$ but how do I show this formally with power series?

Answer 1

Suppose that $y(x)=\sum\limits_{n=0}^\infty a_nx^n$. Plug this into the ODE. Differentiate term-by-term, then group together terms to re-write it as one power series. What do you get?

Answer 2

First assume that $y'=y$ has a power series solution.

Put $y=\sum_{n=0}^{\infty} c_n x^n$, and then $y'=\sum_{n=1}^{\infty} nc_n x^{n-1}$

ie $\sum_{n=1}^{\infty} nc_n x^{n-1}-\sum_{n=0}^{\infty} c_n x^n=\sum_{n=0}^{\infty} [(n+1)c_{n+1} x^{n}-c_n x^n]$, and which gives $c_{n+1}=\frac{c_n}{n+1}$. So $c_0=c_1$, $c_2=\frac{c_0}{2!}$,... $c_k=\frac{c_0}{k!}$.

So $y=c_o e^x$.

be aware that $e^{-x}$ doesn't satisfy, $y'=y$

Answer 3

To begin with, we'd need to suppose that $$y=\sum_{n=0}^\infty a_nx^n$$ for some sequence $\langle a_n\rangle_{n=0}^\infty.$

Next, we'd use the fact that we can differentiate a series term by term to see that \begin{eqnarray}y' &=& \frac{d}{dx}\left[\sum_{n=0}^\infty a_nx^n\right]\\&=& \sum_{n=0}^\infty \frac{d}{dx}\left[a_nx^n\right]\\&=& \sum_{n=0}^\infty a_n\frac{d}{dx}\left[x^n\right]\\&=& \sum_{n=0}^\infty a_n\cdot nx^{n-1}\\&=& \sum_{n=1}^\infty a_n\cdot nx^{n-1}\\&=& \sum_{n=0}^\infty a_{n+1}\cdot(n+1)x^n\end{eqnarray}

After that, we'd see that $$0=y'-y=\sum_{n=0}^\infty\bigl((n+1)a_{n+1}-a_n\bigr)x^n$$ for all $x$ by assumption, letting us conclude that $$(n+1)a_{n+1}-a_n=0$$ for all $n\ge 0.$ Thus, $$a_{n+1}=\frac{a_n}{n+1}$$ for all $n\ge 0,$ so we can readily prove by recursion that $$a_n=\frac{a_0}{n!}$$ for all $n\ge 0.$ Thus, $$y=\sum_{n=0}^\infty\frac{a_0}{n!}x^n=a_0\sum_{n=0}^\infty\frac{x^n}{n!}=a_0e^x.$$