1st order differential equation

[smithg86]

Homework Statement

(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

<br /> y&#039;-y=e^x<br />

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

Homework Equations

e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

The Attempt at a Solution

I plugged y into the DE:
<br /> \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!}<br />

Then I made all series start at n=0:
<br /> \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<br />

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

Set coefficients equal to zero:
<br /> a_{n+1} (n+1) - a_n - \frac{1}{n!} = 0<br />

Solving for recurrence relation:
<br /> a_{n+1} = \frac{n!a_n +1}{(n+1)!}<br />

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

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

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

 

[HallsofIvy]

Homework Statement

(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

<br /> y&#039;-y=e^x<br />

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

Homework Equations

e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

The Attempt at a Solution

I plugged y into the DE:
<br /> \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!}<br />

Then I made all series start at n=0:
<br /> \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<br />

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

Set coefficients equal to zero:
<br /> a_{n+1} (n+1) - a_n - \frac{1}{n!} = 0<br />

Solving for recurrence relation:
<br /> a_{n+1} = \frac{n!a_n +1}{(n+1)!}<br />After plugging in n=0,1,2,3,4... the pattern is:
<br /> a_n = \frac{a_0 + n}{n!}<br />

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

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

You've done remarkably well so far. The standard way to determine the radius of convergence of a power series is to use the "ratio" test.
If a_n= (a_0+ n)/n!, then a_{n+1}= (a_0+ n+1)/(n+1)!. The ratio a_{n+1}x^{n+1}/a_nx^n is
\frac{a_0+ n+ 1}{(n+1)!}\frac{n!}{a_0+ n}x= \frac{a_0+ n+ 1}{n(a_0+ n)}x
what is the limit of that as n goes to infinity? What values of x make it less than 1?

 

[Dick]

You can easily check your result by realizing that the differential equation is pretty easy to solve. It's y=x*e^x+a0*e^x. Expand this and compare with your solution. For radius of convergence, how about trying a ratio test?

 

[smithg86]

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.

 

[Dick]

You can't say y has a infinite radius of convergence until you know what y is. x*y'+y=0 has a solution 1/x which is not going to have infinite radius of convergence solutions - even though all coefficients do. And yes, you caught me. I guessed. The homogeneous part is easy enough (it's linear with constant coefficents) - and I always try guessing the inhomogeneous part before going to more general solutions like variation of parameters.

 

[HallsofIvy]

I'm sorry, Dick. Where did you get xy'+ y = 0 ? The given equation is
y'- y= e^x which has soluion, as you said, y(x)= Ce^x+ xe^x. Wriitten as a power series, that has, of course, radius of convergence infinity- which was what smithg86 said.

 

[smithg86]

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 x_0 is an ordinary point of the differential equation:
P(x)y&#039;&#039; + Q(x)y&#039; + R(x)y = 0
that is, if p=Q/P and q = R/P are analytic at x_0, then the general solution of the differential equation is y = (summation that I don't want to write out) = a_0 y_1 (x) + a_1 y_2 (x). The radius of convergence for each of the series solutions y_1 and y_2 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 x_0?

 

[Dick]

Yes, it was just intended as a counterexample. I don't think it's special because it's first order. It's just because the leading coefficient is vanishing. It does make good sense that if the leading function is nonzero and the subleading functions don't have singularities in some radius R then you can trust the solution out to that radius. Just think about integrating it numerically. So you don't necessarily have to know y before you can conclude something about it's radius of convergence. I stand corrected.