Finding general solution to a differential equation using power series

In summary: This is a summation over an infinite interval and so it's not really a power series.In summary, the student attempted to solve for y near x = 0 of y'' - xy' + 2y = 0 using a power series, but got incorrect results because of the signs of the first terms in the series and because the zero values did not follow the pattern. Furthermore, the student was unable to solve for y using an explicit equation because there are zeros for every even subscript from a_4 onward.
  • #1
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Homework Statement


Find the general solution near x = 0 of y'' - xy' + 2y = 0 (using power series).

Answer:
y = a_0 * y_1(x) + a_1 * y_2(x)
where
y_1(x) = 1 - x^2 and y_2(x) = x - 1/6 * x^3 - 1/120 x^5 - 1/1680 * x^7 - ...


Homework Equations


Power series. Sigma notation for summations. Polynomial derivatives.


The Attempt at a Solution


My attempt is attached. I get y(x) = ke^x and I think I'm wrong but I'm not sure since the answer my book gives doesn't use sigma notation.

I'd appreciate if someone could tell me if I'm right or wrong and if I am wrong, where I went wrong.
Thanks in advance!
 

Attachments

  • MyWork.jpg
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  • #2
I see two mistakes. First, the non-zero an of the second solution aren't given by -a1/n! for n>1. Second, even if that were the case, your solution wouldn't be ex because the sign of the first term the series is wrong.

Try calculating a7 explicitly using the recurrence relation.
 
  • #3
I get a_7 = 1/14 * a_5 = 1/14 * -1/120 * a_1 = -1/1680 * a_1 and I see that the a_n = -1/n! a_1 does not work out for that particular case.

I also see that a_n = -1/n! does not hold for the zero values.

As for the sign of the first term for the e^x part, I didn't understand what you meant. Could you rephrase that please?

I would appreciate it if you could tell me what the explicit relationship is and what the reasoning was in order to get it because my book doesn't cover the material in this way and uses another method which I dislike (the no summation notation, "dot dot dot" method).
 
  • #4
I see more mistakes now that I'm awake. They're easy to see if you just compare the two series. Your solution was
$$y_2(x) = a_1\left(x - \frac{x^3}{6} - \frac{x^5}{120} - \frac{x^7}{1680} - \cdots\right)$$ but
$$-a_1 e^x = a_1\left(-1 - x - \frac{x^2}{2} - \frac{x^3}{6} - \cdots \right)$$ What I meant about the sign of the "first" term was the sign of the linear term.
 
  • #5
Okay, I also redid my work from scratch with neater handwriting and computed up to the a_9 term but I still can't figure out how to get an explicit equation. Namely, how would I take into account that there are zeros for every even subscript from a_4 onward?

I'm currently thinking of something like a_n = [(-1)^n - 1]/2 * 3/n! which works for a_7 and a_9 but not for a_5. I'm still really stuck :(.
 

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  • #6
Why are you insisting on having a closed form expression? Sometimes it's not really possible. Just try writing out more terms and see if you can spot a pattern. It usually doesn't help to simplify as you go because you're looking for a pattern.

Anyway, to get only odd powers, you typically write something like
$$\sin x = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$
 
Last edited:

Related to Finding general solution to a differential equation using power series

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between the rate of change of a dependent variable and one or more independent variables.

2. What is a power series?

A power series is an infinite series in which each term is a multiple of a variable raised to an integer power. It is often used to approximate functions or solve differential equations.

3. How does using a power series help in finding the general solution to a differential equation?

Using a power series allows us to express the solution to a differential equation as a sum of infinitely many terms, making it easier to find a general solution that satisfies all possible initial conditions.

4. What are the steps involved in finding the general solution to a differential equation using power series?

The steps involved include:

  • 1. Write the differential equation in the form of a power series.
  • 2. Substitute the power series into the differential equation.
  • 3. Equate the coefficients of like powers of the variable.
  • 4. Solve for the coefficients using algebraic manipulation.
  • 5. Combine the coefficients to form the general solution.

5. What are some common applications of finding the general solution to a differential equation using power series?

Some common applications include solving problems in physics, engineering, and economics that involve rates of change, growth, or decay. It can also be used to approximate solutions to nonlinear differential equations that cannot be solved analytically.

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