1. ## power series

solving the differential equations

1. y'' - 2xy' + 3y = 0

2. (1+2x)y'' - 3xy' + 2y = 0

solving the differential equations by using method of Frobenius

3. 3x^2 y'' + 2xy' + x^2 y = 0

4. 3xy'' + 2y' + 2y = 0

Is Frobenius one of the crazy mathematicians ?

2. Originally Posted by ggw
y'' - 2xy' + 3y = 0
Let $y(x) = \sum_{n = 0}^{\infty}a_nx^n$

Then
$y^{\prime}(x) = \sum_{n = 1}^{\infty}na_nx^{n - 1}$

$y^{\prime \prime}(x) = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

Substituting these into the differential equation:
$\sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - 2x \cdot \sum_{n = 1}^{\infty}na_nx^{n - 1} + 3 \cdot
\sum_{n = 0}^{\infty}a_nx^n = 0$

$\sum_{n = 2}^{\infty}[n(n - 1)a_n]x^{n - 2} + \sum_{n = 1}^{\infty}[-2na_n]x^n + \sum_{n = 0}^{\infty}[3a_n]x^n = 0$

We need to match powers of x in order to write this as a single sum. Note that the last two sums already have the same power of x listed, so I will write the n = 0 term of the third summation separately, then add the last sums together:
$\sum_{n = 2}^{\infty}[n(n - 1)a_n]x^{n - 2} + \sum_{n = 1}^{\infty}[-2na_n]x^n + 3a_0 + \sum_{n = 1}^{\infty}[3a_n]x^n = 0$

$\sum_{n = 2}^{\infty}[n(n - 1)a_n]x^{n - 2} + \sum_{n = 1}^{\infty}[-2na_n + 3a_n]x^n + 3a_0 = 0$

Now for a trick. The first step is to set $k = n - 2$ in the first summation. This means the sum goes from $k = 2 - 2 = 0$ to $\infty$. Since $n = k + 2$ we get:
$\sum_{k = 0}^{\infty}[(k + 2)(k + 1)a_{k + 2}]x^k + \sum_{n = 1}^{\infty}[-2na_n + 3a_n]x^n + 3a_0 = 0$

So far so good. Now for the last step of the trick: k is just a dummy variable, with no real meaning. So now let's substitute $n = k$:
$\sum_{n = 0}^{\infty}[(n + 2)(n + 1)a_{n + 2}]x^n + \sum_{n = 1}^{\infty}[-2na_n + 3a_n]x^n + 3a_0 = 0$

Now we can write the n = 0 term of the first summation then add the two sums together:
$(2)(1)a_2 + \sum_{n = 1}^{\infty}[(n + 2)(n + 1)a_{n + 2}]x^n + \sum_{n = 1}^{\infty}[-2na_n + 3a_n]x^n + 3a_0 = 0$

$\sum_{n = 1}^{\infty}[(n + 2)(n + 1)a_{n + 2} - 2na_n + 3a_n]x^n + 2a_2 + 3a_0 = 0$

Now, if this expression is going to be true for all x, then all coefficients of x must be 0. Thus
$2a_2 + 3a_0 = 0$

$(n + 2)(n + 1)a_{n + 2} + (3 - 2n)a_n = 0$ for $n \geq 1$

These relations give the values of the coefficients of x when you have initial conditions. As you have not given any initial conditions to go with this equation, I can't go any further for a solution for you.

-Dan

3. Originally Posted by ggw
solving the differential equations

1. y'' - 2xy' + 3y = 0
$\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} - \sum_{n=1}^{\infty} 2na_nx^n + 3 \sum_{n=0}^{\infty} 3a_nx^n = 0$

$\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n - \sum_{n=1}^{\infty} 2na_nx^n + \sum_{n=0}^{\infty} 3a_nx^n=0$

$(2a_2 + 3a_0) + \sum_{n=0}^{\infty}[ (n+2)(n+1)a_{n+2} - 2na_n + 3a_n ] x^n = 0$

Thus,
$1) \ \ \ 3a_0 = - 2a_0$
$2) \ \ \ (n+2)(n+1)a_{n+2} - 2na_n + 3a_n = 0 \mbox{ for }n\geq 1$

The rest is up to you.

Is Frobenius one of the crazy mathematicians ?
I do not know much about him. The only strange thing is that Georg Frobenius was an Algebraist not an Analyst. His method to solve Regular Singular differencial equations was an analysis thing, which is supprising to me.

4. As I learn is different way as you did, but it is the same answer as far as I checked mine.

As I learn, after substitude m into k variables, just solve for a_k+2

then, let k = 1, 2, 3, 4, 5, ,6, 7, ......so on till you see the pattern. then

the solutions will be

y_1 = C_o[1+x^2......]
y_2 = C_1[x - ......]

5. Originally Posted by ThePerfectHacker
Thus,
$1) \ \ \ 2a_2 = - 3a_2$
$2) \ \ \ (n+2)(n+1)a_{n+2} - 2na_n + 3a_n = 0 \mbox{ for }n\geq 1$
Let me actually finish the solution.

1) So we have that $a_2 = - \frac{3}{2}a_0$

2) Here we have $a_{n+2} = \frac{2n-3}{(n+2)(n+1)}\cdot a_n \mbox{ for }n\geq 1$

That means,

$a_4 = \frac{1}{4\cdot 3}\cdot a_2 = - \frac{1}{8}a_0$

$a_6 = \frac{5}{6\cdot 5}\cdot a_4 = - \frac{1}{48}a_0$

Now do the odd terms with $a_1$ being arbitrary.

$a_3 = \frac{-1}{3\cdot 2}a_1 = - \frac{1}{6}a_1$

$a_5 = \frac{5}{5\cdot 4}\cdot a_3 = - \frac{1}{24}a_1$

Thus,
$y = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+...$

Substitute,
$y= a_0+a_1x- \frac{3}{2}a_0x^2 - \frac{1}{6}a_1x^3 - \frac{1}{8}a_0x^4 - \frac{1}{24}a_1 x^5 - \frac{1}{48}a_0 x^6-...$

Rewrite,
$y= a_0\left( 1 - \frac{3}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{48}x^6 - ... \right) + a_1\left( x - \frac{1}{6}x^3 - \frac{1}{24}x^5 - ... \right)$

Thus, the two linearly solutions are given by,
$y_1 = 1 - \frac{3}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{48}x^6 - ...$
$y_2 = x - \frac{1}{6}x^3 - \frac{1}{24}x^5 - ...$

6. thanks a lot, that's much clear to me. I got that...! how about using the # 3&4 how would you express those. thanks a lot.

7. Originally Posted by ggw
3. 3x^2 y'' + 2xy' + x^2 y = 0
Say we are solving for $x>0$.

If we divide through by $3x^2$:
$y'' + \frac{2}{3x}y'+\frac{1}{3}y=0$

We see that if $p(x) = \frac{2}{3x} \mbox{ and }q(x) = \frac{1}{3}$ then does not exist a power series centered at $x=0$ that will express $p(x)$ and $q(x)$ an infinite series. Thus, $x=0$ is a singular point.

However, $\lim_{x\to 0}xp(x) = \frac{2}{3} < \infty \mbox{ and }\lim_{x\to 0}x^2q(x) = 0< \infty$. That means $x=0$ is a regular singular point.

This suggests that we can use the Frobenius Method to solve this differencial equation. The charachteristic equations is: $k(k-1) + \frac{2}{3}k + 0 = 0$.
As obtained from finding those limits above. Thus, $k=0 \mbox{ and }k=\frac{1}{3}$. And note that the roots do not differ by an integer. Which means that we can easily (but with much computation) obtain the two linearly independent solutions.

The above discussion shows that we look for solutions of the form,
$x^k \sum_{n=0}^{\infty}a_n x^n$
Where $k=0,1/3$.

Thus, there are two power series to consider:
$1) \ \ \ \sum_{n=0}^{\infty} a_n x^n$
$2) \ \ \ \sum_{n=0}^{\infty} a_n x^{n+1/3}$

Now substitute that into the differencial equation to get the solutions. That part I leave to you to do, unless you cannot and need me to do it.

8. sum_{n=0}^{infty} a_n x^(n+r)

9. Originally Posted by ggw
sum_{n=0}^{infty} a_n x^(n+r)
yes, but you want to choose the higher root of the indicial equation, which in this case is 1/3

so do the same thing that TPH and topsquark did with the first 2 questions using $\sum_{n=0}^{ \infty} a_n x^{n + 1/3}$

Note that even when we differentiate this sum, the sum still starts at n = 0, don't skip to n = 1, for the first derivative and n = 2 for the second

10. Originally Posted by Jhevon
yes, but you want to choose the higher root of the indicial equation, which in this case is 1/3
There is nothing special about choosing the higher root. That is just what the our textbook does.

11. by frobenius method,
my notes say:

y = \sum_{n=0}^{infty} a_n x^(n+r)

y' = \sum_{n=0}^{infty} (n+r) * a_n * x^(n+r)

y'' = \sum_{n=0}^{infty} (n+r) * (n+r-1) * a_n * x^(n+r-2)

like the first one you did, to solve for C_k+1 =.........

it needs to find two values for r, like the first one you did a_2 = a_0.
this case has two solutions y1 & y2 from the two values of r.

this problem is a little bit different from the first two.

i think after this point, I can do the rest. Thanks

12. Originally Posted by ggw
3. 3x^2 y'' + 2xy' + x^2 y = 0
Look for solution of form: $y=\sum_{n=0}^{\infty}a_n x^{n+1/3}$

Then,
$\sum_{n=0}^{\infty}3(n+1/3)(n-2/3)a_nx^{n+1/3} + \sum_{n=0}^{\infty} 2(n+1/3)a_nx^{n+1/3} + \sum_{n=0}^{\infty} a_nx^{n+1/3 + 2}=0$

Thus,
$\sum_{n=0}^{\infty}3(n+1/3)(n-2/3)a_nx^{n+1/3} + \sum_{n=0}^{\infty} 2(n+1/3)a_nx^{n+1/3} + \sum_{n=2}^{\infty} a_{n-2}x^{n+1/3 }=0$

Evaluate first and second sum at $n=0,1$:
$\left\{ 3\left(\frac{1}{3}\right)\left(-\frac{2}{2}\right)a_0 + 2\left(\frac{1}{3}\right)a_0\right\}x^{1/3} + \left\{3\left(\frac{4}{3}\right) \left(\frac{1}{3}\right)a_1 + 2\left(\frac{4}{3}\right)a_1 \right\}x^{1+1/3}$+ $\sum_{n=2}^{\infty}[3(n+1/3)(n-2/3)a_n+2(n+1/3)a_n+a_{n-2}]x^{n+1/3}=0$

The first sum (enclosed by { } ) is equal to zero by our special choice of $k=1/3$.

The second sum has to be zero and the only way that that can happen is if $a_1=0$.

What remains is the summation need to be zero thus:
$\sum_{n=2}^{\infty}[3(n+1/3)(n-2/3)a_n+2(n+1/3)a_n+a_{n-2}]x^{n+1/3}=0$

Thus, the recurrence relation is given by:
$3(n+1/3)(n-2/3)a_n+2(n+1/3)a_n+a_{n-2} = 0 \mbox{ for }n\geq 0$

~~~
Off Topic

At this point I should skip steps (because you are expected to do the algebra) but since I do not want you to complain about the algebra I will tell you everything you need to know.

Move the smaller term over to the right hand side:
$3(n+1/3)(n-2/3)a_n + 2(n+1/3)a_n = -a_{n-2}$
Divide, by $(n+1/3)$ both sides to get:
$3(n-2/3)a_n+2a_n = - \frac{a_{n-2}}{(n+1/3)}$
Combine the left hand side,
$(3n - 2)a_n + 2a_n = - \frac{a_{n-2}}{(n+1/3)}$
Combine the left hand side again,
$3na_n = - \frac{a_{n-2}}{(n+1/3)}$
Divide by coefficient in front,
$a_n = - \frac{a_{n-2}}{3n(n+1/3)}$
Multiply the three inside the paranthesis on right hand,
$\boxed{a_n = - \frac{a_{n-2}}{n(3n+1)} \mbox{ for }n\geq 2 }$

Back on Topic
~~~
Remember we found that $a_1=0$ the above recurrent relation is defined by difference of two's that means $a_3=a_5=a_7=...0$.

Now let $a_0$ be an arbitrary non-zero real number.
Then,
$a_2 = - \frac{a_0}{2\cdot 7}$

$a_4 = - \frac{a_2}{4\cdot 13} = \frac{a_0}{2\cdot 4 \cdot 7\cdot 13}$

$a_6 = - \frac{a_4}{6\cdot 19} = - \frac{a_0}{2\cdot 4\cdot 6 \cdot 7\cdot 13 \cdot 19}$

So in general,
$a_{2n} = \frac{(-1)^na_0}{(2\cdot 4\cdot 6\cdot ... \cdot 2n)(7\cdot 13\cdot ... \cdot (3n+1))} = \frac{(-1)^na_0}{2^n \cdot n! \cdot (7\cdot .... \cdot (3n+1))}$

That means the solution is given by modified power series,
$y = a_0 \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1/3}}{2^n \cdot n! \cdot (7\cdot .... \cdot (3n+1)) }$

There is still a second solution!

13. Originally Posted by ggw
3. 3x^2 y'' + 2xy' + x^2 y = 0
Now we look for the solution of the form with $k=0$:
$y=\sum_{n=0}^{\infty} a_n x^n$

This one is going to be simpler, start by doing the usual,

$\sum_{n=0}^{\infty}3(n+2)(n+1)a_{n+2}x^{n+2} + \sum_{n=0}^{\infty} 2(n+1)a_{n+1}x^{n+1} + \sum_{n=0}^{\infty}a_nx^{n+2}=0$

Change index on first and third:

$\sum_{n=1}^{\infty}3(n+1)na_{n+1}x^{n+1} + \sum_{n=0}^{\infty} 2(n+1)a_{n+1}x^{n+1} + \sum_{n=1}^{\infty}a_{n-1}x^{n+1}=0$

Evaluate the second at $x=0$ to get:
$2a_1 x + \sum_{n=1}^{\infty}[3(n+1)na_{n+1}+2(n+1)a_{n+1}+a_{n-1}]x^{n+1} = 0$

This tells us that $a_1=0$
And the recurrence relations:
$3(n+1)na_{n+1} + 2(n+1)a_{n+1}+a_{n-1} = 0 \mbox{ for }n\geq 1$
Divide by $n+1$ to get:
$(3n+2)a_{n+1} = -\frac{a_{n-1}}{(n+1)}$
Divide again,
$a_{n+1} = - \frac{a_{n-1}}{(3n+2)(n+1)}$
Since $a_1=0$ the recurrence relation tells us that $0=a_3=a_5=a_7=...$

Let $a_0$ be arbitrary non-zero real number then,
$a_2 = - \frac{a_0}{5\cdot 2}$

$a_4 = - \frac{a_2}{11\cdot 4} = \frac{a_0}{(5\cdot 11)\cdot (2\cdot 4)}$

$a_6 = -\frac{a_4}{17 \cdot 6 } = - \frac{a_0}{(5\cdot 11\cdot 17)\cdot (2\cdot 4\cdot 6)}$

In general we have that,

$a_{2n} = \frac{(-1)^na_0}{[5\cdot 11\cdot ... \cdot (6n-1)]\cdot (2\cdot 4\cdot ... 2n)} = \frac{(-1)^n a_0}{2^n \cdot n![5\cdot 11\cdot ... \cdot (6n-1)]}$

So the solution is given by,
$y= a_0 \sum_{n=0}^{\infty}\frac{(-1)^n x^n}{2^n \cdot n![5\cdot 11\cdot ... \cdot (6n-1)]}$

14. Originally Posted by ThePerfectHacker
There is nothing special about choosing the higher root. That is just what the our textbook does.
why do they do it then, they must have some reasoning behind it

15. Originally Posted by Jhevon
why do they do it then, they must have some reasoning behind it
Because the solutions for $k$ can be:
1)Two real solutions and do not differ by an integer.
2)Two real solutions that differ by an integer.
3)Only one real solution.
4)Two complex solutions.

We did only #1, that is by far the easiet. The problem with #2 is that you get two linearly dependent solutions because the infinite series keeps increasing in terms of integer powers. The problem with #3 is that you need another one. And the problem with #4 is that they are not real. Each of cases: 2,3,4 is much more difficult. You can find their methods in a book on differencial equations. That is why they just tell you to use the larger one just in case you reach that problem.