Simplification step: solving a diff eqn using a power series

[Sparky_]

Homework Statement

Hello,

I suspect this is an easy answer but I am not seeing it. I am reviewing (more so for fun / hobby) some differential equations – I’m not in school.

I’m needing help with an example problem in Differential Equations With Boundary-Value Problems Zill 2nd edition. In googling around for a possible errata, I see it’s also in other editions – not an error – it’s me.

It’s Chapter 6, section 2, example7, of the 2nd edition (my book from way back) and Chapter 6, section 1, example 4 of the 7th edition.

The problem is to solve by way of a power series:

(x² - 1 ) y'' + x y - y = 0

Homework Equations

The Attempt at a Solution

(again) - the solution is in the book - it's an example problem

y = ∑c_nxⁿ
y' = ∑ (n) c_nx^n-1
y'' = ∑ (n) (n-2) c_nx^n-2

(x² +1 ) ∑(n)(n-1)c_nx^n-2 + (x)∑(n)c_nx^n-1 - ∑c_nxⁿ
Divided through by x² (after cross multiplying the x² + 1)

the first summation the x²'s cancel, the second summation (the term that results from multiplying by the "1" will subtract the "2" exponent in the x^n-2 term, the third summation leaves an x in the denominator and as a result will subtract a "1" in the exponent of x^n-1= ∑ (n)(n-1) c_n xⁿ + ∑ (n)(n-1)c_n x^n-2 + ∑(n)c_n xⁿnow for my question - the forth summation - the "-y" term in the original equation

dividing through by x² - it looks like the last summation should be ∑c_n x^-2

the book has ∑c_n xⁿ for the last term - as if it was not divided by the x² term

??
what am I missing, regarding the last term ( the lone y term) after dividing through by x²?

Thanks

 

[Ray Vickson]

Homework Statement

Hello,

I suspect this is an easy answer but I am not seeing it. I am reviewing (more so for fun / hobby) some differential equations – I’m not in school.

I’m needing help with an example problem in Differential Equations With Boundary-Value Problems Zill 2nd edition. In googling around for a possible errata, I see it’s also in other editions – not an error – it’s me.

It’s Chapter 6, section 2, example7, of the 2nd edition (my book from way back) and Chapter 6, section 1, example 4 of the 7th edition.

The problem is to solve by way of a power series:

(x² - 1 ) y'' + x y - y = 0
[/B]

Homework Equations

The Attempt at a Solution

(again) - the solution is in the book - it's an example problem

y = ∑c_nxⁿ
y' = ∑ (n) c_nx^n-1
y'' = ∑ (n) (n-2) c_nx^n-2

(x² +1 ) ∑(n)(n-1)c_nx^n-2 + (x)∑(n)c_nx^n-1 - ∑c_nxⁿ
Divided through by x² (after cross multiplying the x² + 1)

the first summation the x²'s cancel, the second summation (the term that results from multiplying by the "1" will subtract the "2" exponent in the x^n-2 term, the third summation leaves an x in the denominator and as a result will subtract a "1" in the exponent of x^n-1= ∑ (n)(n-1) c_n xⁿ + ∑ (n)(n-1)c_n x^n-2 + ∑(n)c_n xⁿ


now for my question - the forth summation - the "-y" term in the original equation

dividing through by x² - it looks like the last summation should be ∑c_n x^-2

the book has ∑c_n xⁿ for the last term - as if it was not divided by the x² term

??
what am I missing, regarding the last term ( the lone y term) after dividing through by x²?

Thanks


[/B]

Please turn off the bold font: it is distracting.

You wrote an $x^2-1$ coefficient in the original DE but an $x^2+1$ coefficient in the series. Which one should it be?

Also: life would be much simpler for you if you switched from typing with "SUB" and "SUP" constructs to plain LaTeX. For example, it is a lot easier to type x^2 or c_n than x[ S U P]2[/ S U P] or c[S U B] n [/S U B] (where I have inserted spaces to prevent the parser from making an actual superscript or subscript). However, LaTex constructs must appear between ## to start and end; that is, we would type # # x^2 # # (again, with spaces inserted to prevent the parser from actually implementing LaTeX on the expression).

 

[Sparky_]

Please turn off the bold font: it is distracting.

You wrote an $x^2-1$ coefficient in the original DE but an $x^2+1$ coefficient in the series. Which one should it be?

Also: life would be much simpler for you if you switched from typing with "SUB" and "SUP" constructs to plain LaTeX. For example, it is a lot easier to type x^2 or c_n than x[ S U P]2[/ S U P] or c[S U B] n [/S U B] (where I have inserted spaces to prevent the parser from making an actual superscript or subscript). However, LaTex constructs must appear between ## to start and end; that is, we would type # # x^2 # # (again, with spaces inserted to prevent the parser from actually implementing LaTeX on the expression).

it's (x^2 + 1), regarding the other - i was using the menu at the top of the window for sub and super script, I have no idea how the bold was turned on

 

[Mark44]

Please turn off the bold font: it is distracting.

I undid the bold font style in the OP.

I have no idea how the bold was turned on

It's as a result of typing immediately after the bolded headings. I'll look into getting that fixed, since this is probably that occurs often.

 

[Ray Vickson]

it's (x^2 + 1), regarding the other - i was using the menu at the top of the window for sub and super script, I have no idea how the bold was turned on

I think it happens if you type your results inside the "[B O L D] ... [/ B O L D ]" pair embodied in the PF headings. If you just type outside that environment, you will get a standard font. However, you can still make some of your inputs bold for the sake of emphasis.

Anyway, your method of trying to divide by $x$ and/or $x^2$ is something you should absolutely NEVER do: it is a fatal error. Why? Well, you are looking for a series solution near $x = 0$, so dividing by $x$ is a violation. Anyway, it is not necessary.

Here is how I used to do it back in the Stone Age when I was taking this material. I always found it useful to write out explicitly the first 3 or 4 terms before switching to summation notation. So, if $y = c_0+ c_1 x + c_2 x^2 + c_3 x^3 + \cdots,$ we have
$$
\begin{array}{rcl}
y'&=&c_1 + 2 c_2 x + 3 c_3 x^2 + \cdots , \; \text{hence}\\
x y'&=& c_1 x + 2 c_2 x^2 + 3 c_3 x^3 + \cdots = c_1 x + \sum_{n=2}^{\infty} n c_n x^n \\
y''&=& 2 \cdot 1 c_2 + 3 \cdot 2 c_3 x + 4 \cdot 3 c_4 x^2 + \cdots = 2 c_2 + 6 c_3 x + \sum_{n=2}^{\infty} (n+2)(n+1) c_{n+2} x^n \\
x^2 y'' &=& 2 \cdot 1 c_2 x^2 + 3 \cdot 2 c_3 x^3 + \cdots = \sum_{n=2}^{\infty} n(n-1) c_n x^n
\end{array}
$$
Thus, the left-hand-side of the DE is
$$-c_0+ 2 c_2 +(c_1 - c_1+6 c_3) x + \sum_{n=2}^{\infty} [n(n-1)c_n - c_n + n c_n+(n+2)(n+1) c_{n+2} ] x^n$$
This is supposed to vanish for all $x$, so it give you some relationships between the coefficients $c_j$.

 

[vela]

(x² +1 ) ∑(n)(n-1)c_nx^n-2 + (x)∑(n)c_nx^n-1 - ∑c_nxⁿ
Divided through by x² (after cross multiplying the x² + 1)

I'm not sure what you mean by "cross multiplying" in this context, and as far as I can tell, you didn't divide through by $x^2$. What's happening with the y'' term is the following:

\begin{align*}
(x^2+1)\sum n(n-1)c_nx^{n-2} &= \sum n(n-1)c_n(x^2+1)x^{n-2} \\
&= \sum n(n-1)c_n(x^n+x^{n-2}) \\
&= \sum n(n-1)c_n x^n + \sum n(n-1)c_n x^{n-2}
\end{align*}
Same idea with the y' term. There's no coefficient in front of the y term, however, so the summation remains unchanged.

 

[Sparky_]

I think it happens if you type your results inside the "[B O L D] ... [/ B O L D ]" pair embodied in the PF headings. If you just type outside that environment, you will get a standard font. However, you can still make some of your inputs bold for the sake of emphasis.

Anyway, your method of trying to divide by $x$ and/or $x^2$ is something you should absolutely NEVER do: it is a fatal error. Why? Well, you are looking for a series solution near $x = 0$, so dividing by $x$ is a violation. Anyway, it is not necessary.

Here is how I used to do it back in the Stone Age when I was taking this material. I always found it useful to write out explicitly the first 3 or 4 terms before switching to summation notation. So, if $y = c_0+ c_1 x + c_2 x^2 + c_3 x^3 + \cdots,$ we have
$$
\begin{array}{rcl}
y'&=&c_1 + 2 c_2 x + 3 c_3 x^2 + \cdots , \; \text{hence}\\
x y'&=& c_1 x + 2 c_2 x^2 + 3 c_3 x^3 + \cdots = c_1 x + \sum_{n=2}^{\infty} n c_n x^n \\
y''&=& 2 \cdot 1 c_2 + 3 \cdot 2 c_3 x + 4 \cdot 3 c_4 x^2 + \cdots = 2 c_2 + 6 c_3 x + \sum_{n=2}^{\infty} (n+2)(n+1) c_{n+2} x^n \\
x^2 y'' &=& 2 \cdot 1 c_2 x^2 + 3 \cdot 2 c_3 x^3 + \cdots = \sum_{n=2}^{\infty} n(n-1) c_n x^n
\end{array}
$$
Thus, the left-hand-side of the DE is
$$-c_0+ 2 c_2 +(c_1 - c_1+6 c_3) x + \sum_{n=2}^{\infty} [n(n-1)c_n - c_n + n c_n+(n+2)(n+1) c_{n+2} ] x^n$$
This is supposed to vanish for all $x$, so it give you some relationships between the coefficients $c_j$.

Thank you - i repeated your suggestion and things worked out. I still believe I can get there (minus avoiding x = 0 in the denominator) with my original path but I'm not going to dwell on it.

Thank you so much!

 

[Sparky_]

I'm not sure what you mean by "cross multiplying" in this context, and as far as I can tell, you didn't divide through by $x^2$. What's happening with the y'' term is the following:

\begin{align*}
(x^2+1)\sum n(n-1)c_nx^{n-2} &= \sum n(n-1)c_n(x^2+1)x^{n-2} \\
&= \sum n(n-1)c_n(x^n+x^{n-2}) \\
&= \sum n(n-1)c_n x^n + \sum n(n-1)c_n x^{n-2}
\end{align*}
Same idea with the y' term. There's no coefficient in front of the y term, however, so the summation remains unchanged.

what I meant by cross multiplying is, I multiplied first the x^2 term and then the "1" ending with an x^2 in front of a summation and also just the summation (the 1 times the summation)

 

[Ray Vickson]

what I meant by cross multiplying is, I multiplied first the x^2 term and then the "1" ending with an x^2 in front of a summation and also just the summation (the 1 times the summation)

Right: that is exactly what you end up with. Then you need to change the index of summation so that aside from the first few terms (which do not fit exactly) the remaining terms all involve terms in the same $x^n$. To keep things straight, I recommend against compressing everything into the smallest-looking formulas, but rather, do what I did: write the first few terms one-by-one and then put summation signs on the rest. That is how it was done in all the books on DEs that I have ever owned.

 

[Sparky_]

Right: that is exactly what you end up with. Then you need to change the index of summation so that aside from the first few terms (which do not fit exactly) the remaining terms all involve terms in the same $x^n$. To keep things straight, I recommend against compressing everything into the smallest-looking formulas, but rather, do what I did: write the first few terms one-by-one and then put summation signs on the rest. That is how it was done in all the books on DEs that I have ever owned.

Thank you (again) !

( I had the first 3 terms but did something stupid on the last one ) when I read your post and repeated things - it worked (agreed)