1st order differential equation

1. The problem statement, all variables and given/known data

(Can someone please check my work? Bear with me, this is my first time using LaTex on this forum...)
Find the general solution to the first oder ODE

[tex]
y'-y=e^x
[/tex]

by substituting a series
[tex]y= \sum_{n=0}^\infty a_n x^n[/tex] about [tex]x_0=0[/tex], finding the recurrence relation for a_n, and solving to find an expression for the general term [tex]a_n[/tex] in terms of [tex]a_0[/tex]. What is the radius of convergence of the solution?

2. Relevant equations

[tex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]

3. The attempt at a solution

I plugged y into the DE:
[tex]
\sum_{n=1}^\infty a_n n x^{n-1} - \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty \frac{x^n}{n!}
[/tex]

Then I made all series start at n=0:
[tex]
\sum_{n=0}^\infty a_{n+1} (n+1) x^n - \sum_{n=0}^\infty a_n x^n - \sum_{n=0}^\infty \frac{x^n}{n!} = 0
[/tex]

Bringing together like terms:
[tex]
\sum_{n=0}^\infty x^n (a_{n+1} (n+1) - a_n - \frac{1}{n!}) = 0
[/tex]

Set coefficients equal to zero:
[tex]
a_{n+1} (n+1) - a_n - \frac{1}{n!} = 0
[/tex]

Solving for recurrence relation:
[tex]
a_{n+1} = \frac{n!a_n +1}{(n+1)!}
[/tex]

After plugging in n=0,1,2,3,4... the pattern is:
[tex]
a_n = \frac{a_0 + n}{n!}
[/tex]

Therefore, the solution is:
[tex]
y = a_0 + (a_0 + 1)x+(\frac{a_0 + 2}{2})x^2 + (\frac{a_0 + 3}{6})x^3 +...
[/tex]

I was unsure about how to determine the radius of convergence.

 

HallsofIvy, the limit of the ratio test as n goes to infinity is zero, so x can be infinite; the radius of convergence is infinite. Could you say the radius of convergence is infinite, as all parts of the DE have infinite radii of convergence? (i.e., 1*y'-1*y=e^x; 1 and e^x have an infinite radii of convergence)?

Dick, what method did you use to solve the differential equation? By looking at the equation, I could've guessed y=x*e^x + c, but I can't figure out how to do it with any of the methods we've learned in my intro to DE class.

 

I think Dick was using xy' + y = 0 as a counterexample to what I said about guessing about the radius of convergence of the solution based on the coefficients of the original equation. I looked in my textbook, and I found something about the radius of convergence for a 2nd order DE. I was trying to do something similar with this 1st order DE. Here's what my textbook says:

"If [tex]x_0[/tex] is an ordinary point of the differential equation:
[tex]P(x)y'' + Q(x)y' + R(x)y = 0[/tex]
that is, if p=Q/P and q = R/P are analytic at [tex]x_0[/tex], then the general solution of the differential equation is y = (summation that I don't want to write out) = [tex]a_0 y_1 (x) + a_1 y_2 (x)[/tex]. The radius of convergence for each of the series solutions [tex]y_1[/tex] and [tex]y_2[/tex] is at least as large as the minimum of the radii of convergence of the series for p and q."

Now that I think about it, does the above property not hold for 1st order DE's because P(x)=0, and therefore p and q are not analytic at [tex]x_0[/tex]?