Diff. EQ: How do I solve a 2nd order linear EQ with series?

[VinnyCee]

This is problem number 1 (yes, one) in chapter 5.2 in Boyce, DiPrima, 8th edition.

y'' - y = 0, x_0 = 0

Substituting the series for y and y-double prime:
\sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} - \sum_{n = 0}^{\infty} a_n x^n = 0

Now, substituting n + 2 for n in first term:
\sum_{n = 0}^{\infty} (n + 2)(n + 1) a_{n + 2} x^n - \sum_{n = 0}^{\infty} a_n x^n = 0

Combining the series and factoring out x^n:
\sum_{n = 0}^{\infty} \left[(n + 2)(n + 1) a_{n + 2} - a_n\right] x^n = 0

Arriving at the recursion formula:
a_{n + 2} = \frac{a_n}{(n + 2)(n + 1)}

This is where I am completely stuck. What is it I am supposed to assume for ao and a1? I know they are arbitrary, but what does that mean and how does it help me to solve the differential equation?

The answer given is:
y_1(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ... = \sum_{n = 0}^{\infty} \frac{x^{2n}}{(2n)!} = cosh x

y_2(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ... = \sum_{n = 0}^{\infty} \frac{x^{2n + 1}}{(2n + 1)!} = sinh x

 

[OlderDan]

This is problem number 1 (yes, one) in chapter 5.2 in Boyce, DiPrima, 8th edition.

y'' - y = 0, x_0 = 0

Substituting the series for y and y-double prime:
\sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} - \sum_{n = 0}^{\infty} a_n x^n = 0

Now, substituting n + 2 for n in first term:
\sum_{n = 0}^{\infty} (n + 2)(n + 1) a_{n + 2} x^n - \sum_{n = 0}^{\infty} a_n x^n = 0

Combining the series and factoring out x^n:
\sum_{n = 0}^{\infty} \left[(n + 2)(n + 1) a_{n + 2} - a_n\right] x^n = 0

Arriving at the recursion formula:
a_{n + 2} = \frac{a_n}{(n + 2)(n + 1)}

This is where I am completely stuck. What is it I am supposed to assume for ao and a1? I know they are arbitrary, but what does that mean and how does it help me to solve the differential equation?

The answer given is:
y_1(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ... = \sum_{n = 0}^{\infty} \frac{x^{2n}}{(2n)!} = cosh x

y_2(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ... = \sum_{n = 0}^{\infty} \frac{x^{2n + 1}}{(2n + 1)!} = sinh x

If you multiplied these solutions by a constant, would they still be solutions? What values have been chosen for a_0 and a_1 to get the solutions in the form they are given. In many problems, there are initial conditions given, which impose specific values on the "arbitrary constants". When no initial conditions are specified, you can choose convenient values and still have a solution.

 

[VinnyCee]

If you multiplied these solutions by a constant, would they still be solutions? What values have been shosen for a_0 and a_1 to get the solutions in the form they are given. In many problems, there are initial conditions given, which impose specific values on the "arbitrary constants". When no initial conditions are specified, you can choose convenient values and still have a solution.

How do I arrive at the solution given? I know that it has something to do with the arbitrary values or something. Like a0 = 0 and a1 = 1. But what is that process after choosing values of a0 and a1?

 

[Pyrrhus]

Notice that you can analyze this recurrence relation through even coefficients and odd coefficients separately, starting by n=0,1,2,3...

 

[dextercioby]

Do you know to solve this equation "the old fashioned way"...?From the general recusion formula,can u get the form of the general term...?

Daniel.

 

[VinnyCee]

Here is the first few for n = 1, 2, 3, 4 ...

a_{n + 2} = \frac{a_n}{(n + 2)(n + 1)}

n = 1: a_{(1) + 2} = \frac{a_{(1)}}{\left[(1) + 2\right] \left[(1) + 1\right]} = \frac{a_1}{3*2} = a_3

n = 2: a_{(2) + 2} = \frac{a_{(2)}}{\left[(2) + 2\right] \left[(2) + 1\right]} = \frac{a_2}{4*3} = a_4

n = 3: a_{(3) + 2} = \frac{a_{(3)}}{\left[(3) + 2\right] \left[(3) + 1\right]} = \frac{a_1}{5*4*3*2} = a_5

n = 4: a_{(4) + 2} = \frac{a_{(4)}}{\left[(4) + 2\right] \left[(4) + 1\right]} = \frac{a_2}{6*5*4*3} = a_6

I know there is a pattern, but where and how do I represent it?

 

[Justin Lazear]

You could start with a₀ = a₀ and a₁ = a₁. Or a₀ = c₁ and a₁ = c₂. Just eliminate as many of the a_n constants as you can.

Your denominator looks awfully similar to a factorial... And how did you get a₄ = a₂ = a₀?

--J

 

[VinnyCee]

Fixed recursions above

How do i solve now? I have a recusion that involves a1 and a2. How do I solve for the a1 and a2? What about a0?

 

[Justin Lazear]

a₂ is when n = 0. Calculate it for n = 0. You must choose two of the constants to be arbitrary. A most convenient choice would be a₀ and a₁.

--J

 

[VinnyCee]

Cool! One step closer...

n = 0: a_{(0) + 2} = \frac{a_{(0)}}{\left[(0) + 2\right] \left[(0) + 1\right]} = \frac{a_0}{2*1} = a_2

n = 1: a_{(1) + 2} = \frac{a_{(1)}}{\left[(1) + 2\right] \left[(1) + 1\right]} = \frac{a_1}{3*2} = a_3

n = 2: a_{(2) + 2} = \frac{a_{(2)}}{\left[(2) + 2\right] \left[(2) + 1\right]} = \frac{a_2}{4*3} = \frac{a_0}{4*3*2*1} = a_4

n = 3: a_{(3) + 2} = \frac{a_{(3)}}{\left[(3) + 2\right] \left[(3) + 1\right]} = \frac{a_1}{5*4*3*2*1} = a_5

n = 4: a_{(4) + 2} = \frac{a_{(4)}}{\left[(4) + 2\right] \left[(4) + 1\right]} = \frac{a_0}{6*5*4*3*2*1} = a_6

Now what?

 

[Justin Lazear]

Now rewrite your factorials in factorial form and deduce an expression for a_2n and a_2n+1 in terms of a₀ and a₁, respectively.

--J

 

[VinnyCee]

n = 0: a_2 = \frac{a_0}{2!}

n = 1: a_3 = \frac{a_1}{3!}

n = 2: a_4 = \frac{a_0}{4!}

n = 3: a_5 = \frac{a_1}{5!}

n = 4: a_6 = \frac{a_0}{6!}

Now I separate(?) and i get this:

y_1(x) = \frac{a_1}{3!} + \frac{a_1}{5!} + \frac{a_1}{7!}

y_2(x) = \frac{a_0}{2!} + \frac{a_0}{4!} + \frac{a_0}{6!}

WHat do I do now? It seems that there are terms missing.

 

[VinnyCee]

Now I have it! Write out the first few terms from the recursion formula:

y(x)\,=\,\sum_{n\,=\,0}^{\infty}\,a_n\,x^n\,=\,a_0\,+\,a_1\,x\,+\,\frac{a_0}{2!}\,x^2\,+\,\frac{a_1}{3!}\,x^3\,+\,\frac{a_0}{4!}\,x^4\,+\,\frac{a_1}{5!}\,x^5\,+\,\frac{a_0}{6!}\,x^6\,+\,...

Now, separate the a0 and a1 terms:

a_0\left[1+\frac{x^2}{2!}\,+\,\frac{x^4}{4!}\,+\,\frac{x^6}{6!}\,+\,...\right]\,+\,a_1\,\left[x\,+\,\frac{x^3}{3!}\,+\,\frac{x^5}{5!}\,+\,\frac{x^7}{7!}\,+\,...\right]\,=\,a_0\,cosh\,x\,+\,a_1\,sinh\,x

The only thing, is how do I know what the a0 and a1 terms are beforehand? I don't. Can someone please tell me why this a0 and a1 assumption can be made? And how it can be made for other general EQ's?

Here is a http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp that tells all about this series stuff for differential equations. That is where I got the a0 and a1 assumptions from.

 

[VinnyCee]

I know now...

The reason for the first a0 and a1 terms being known is the fact that we are looking for an equation in the form of \sum_{n\,=\,0}^{\infty}\,a_n\,x^n,and solving for n = 0 and n = 1 gives these first two terms!

Great stuff thanks for the help people!

 

[Justin Lazear]

You know that there must be two arbitrary constants because the differential equation is of second order. You select the 0 and 1 constants because they are generally the most convenient when you have these problems that are designed to work out nicely. If your differential equation were instead 2(y''-y)=0, you would very likely choose your arbitrary constants to be 2a₀ and 2a₁.

Ultimately, your strategy is to just pick whichever two you feel like it and then change them if you see an advantage to a particular choice, as in this case. Choosing the 0 and 1 constants allows you to write your solutions as a linear combination of the sinh and cosh functions, whereas others would require you to have a linear combination of a multiple of the sinh and cosh functions. It's simply prettier this way, and pretty is good.

--J