I am a bit rusty, can anyone help solve this problem with me?

[Johnny5]

[math] 3x^3 y^{\prime \prime \prime} + 9x y^\prime - y = 0 [/math]

 

Where prime denotes differentiation with respect to x.

 

i.e. y`=dy/dx

 

Thank you

 

Well I will just begin working on my own problem.

 

Assume a power series solution.

 

[math] y(x) = \sum_{n=0}^{n=\infty} C_n x^n [/math]

 

So the first derivative with respect to x is given by:

 

[math] y^\prime = \frac{dy}{dx} = \sum_{n=0}^{n=\infty} nC_n x^{n-1 [/math]

 

The second derivative is given by:

 

[math] y^{\prime \prime} = \frac{d^2y}{dx^2} = \sum_{n=0}^{n=\infty} n(n-1)C_n x^{n-2 [/math]

 

The third derivative is given by:

 

[math] y^{\prime \prime \prime} = \frac{d^3y}{dx^3} = \sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n-3 [/math]

 

Three x cubed times the third derivative is given by:

 

[math] 3x^3 y^{\prime \prime \prime} = 3x^3 \sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n-3 [/math]

 

Nine x times the first derivative is given by:

 

[math] 9x y^\prime = \frac{dy}{dx} = 9x \sum_{n=0}^{n=\infty} nC_n x^{n-1 [/math]

 

And negative one times y is given by:

 

[math] -y = -\sum_{n=0}^{n=\infty} C_n x^n [/math]

 

Adding these three series together yields:

 

[math] 3x^3 y^{\prime \prime \prime} + 9x y^\prime -y = [/math]

 

[math] 3x^3 \sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n-3} + 9x \sum_{n=0}^{n=\infty} nC_n x^{n-1} -\sum_{n=0}^{n=\infty} C_n x^n = 0 [/math]

 

Now, let's perform the necessary indicial operations, so that we can write the LHS as a single series, if possible.

 

The first thing to notice is that we can move x^3 through the series, as well as x, like so:

 

[math] 3\sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n} + 9 \sum_{n=0}^{n=\infty} nC_n x^{n} -\sum_{n=0}^{n=\infty} C_n x^n = 0 [/math]

 

Actually, in this case no indicial operations need to be performed, because we can factor out x^n now.

 

[math] \sum_{n=0}^{n=\infty}

[3n(n-1)(n-2)C_n + 9n C_n - C_n ]x^n = 0 [/math]

 

And we can also factor out C_n like so:

 

[math] \sum_{n=0}^{n=\infty}

C_n[3n(n-1)(n-2) + 9n - 1 ]x^n = 0 [/math]

 

The only way for the statement above to be true, given that not(x=0), is if:

 

[math] C_n [3n(n-1)(n-2) + 9n - 1 ] = 0 [/math]

 

Now, C_n is just an arbitrary constant. Let us stipulate that it be nonzero, so that we can divide through by it. Thus:

 

[math] [3n(n-1)(n-2) + 9n - 1 ] = 0 [/math]

 

Now, let us carry out the multiplication above, and write the LHS as a cubic equation, which will have exactly three roots.

 

[math] (n-1)(n-2) = n^2-2n-n+2 = n^2-3n+2 [/math]

 

[math] 3n(n^2-3n+2 ) = 3n^3-9n^2+6n [/math]

 

 

[math] 3n(n^2-3n+2 )+9n-1 = 3n^3-9n^2+6n +9n-1[/math]

 

Now,

 

[math] 3n(n-1)(n-2) + 9n - 1 = 3n(n^2-3n+2 )+9n-1 [/math]

 

And

 

[math] 3n^3-9n^2+6n +9n-1 = 3n^3-9n^2+15n-1[/math]

 

Therefore, using the transitive property of equality, as well as the previous three lines of work, it follows that:

 

[math] 3n(n-1)(n-2) + 9n - 1 = 3n^3-9n^2+15n-1 [/math]

 

Therefore:

 

If

[math] [3n(n-1)(n-2) + 9n - 1 ] = 0 [/math]

 

then

 

[math] 3n^3-9n^2+15n-1 =0 [/math], and conversely.

 

So now, we need to find the roots of the above polynomial.

 

The method I am going to employ is called synthetic division.

 

Synthetic division

 

This Integer Root Theorem is an instance of the more general Rational Root Theorem: If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

 

Lets look at the factors of the constant term. The constant term is -1.

 

[math] 1 \cdot -1 = -1 [/math]

 

Lets look at the factors of the leading coefficient:

 

[math] 1 \cdot 3 = 3 [/math]

 

So we can check for roots from amongst:

 

[math] \pm 1 [/math]

 

[math] \pm \frac{1}{3} [/math]

 

And we are going to do this using synthetic division.

 

Checking -1:

[math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math]

[math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|-1} [/math]

[math] .\ \ \ -3 \ \ 12 \ \ -27[/math]

__________________

[math] 3 \ \ -12 \ \ 27 \ \ -28 [/math]

 

Because the remainder, -28, isn't zero, -1 is not a root.

 

Checking 1:

 

[math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math]

[math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|1} [/math]

[math] .\ \ \ \ 3 \ \ -6 \ \ \ 9[/math]

__________________

[math] 3 \ \ -6 \ \ 9 \ \ -8 [/math]

 

Because the remainder, -8, isn't zero, 1 is not a root.

 

Checking 1/3:

 

[math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math]

[math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|\frac{1}{3} } [/math]

[math] .\ \ \ \ 1 \ \ \ \frac{-8}{3} \ \frac{37}{9} [/math]

__________________

[math] 3 \ \ -8 \ \ \frac{37}{3} \ \frac{28}{9} [/math]

 

Because the remainder, 28/9, isn't zero, 1/3 is not a root.

 

Checking -1/3:

 

[math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math]

[math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|\frac{-1}{3} } [/math]

[math] .\ \ \ -1 \ \ \ \frac{10}{3} \ \frac{-55}{9} [/math]

__________________

[math] 3 \ \ -10 \ \ \frac{55}{3} \ \frac{-64}{9} [/math]

 

Because the remainder, -64/9, isn't zero, -1/3 is not a root.

 

Therefore, there are no rational roots of the polynomial.

 

But, n is constrained to be a positive integer.

[uncool]

First thing to assume is that y is a simple power function - because each term is in the form x^a*(ath derivative of y). Find all working a's. There should be 3 of them. Let the function be b1*x^a1 + b2*x^a2 + b3*x^a3, where b1, b2, and b3 are any constants. This should solve the equation.

If there are two equal solutions (i.e. a1 = a2), then instead use b1*x^a1+b2*x^a1*ln(x) + b3*x^a3, etc.

-Uncool-

[Johnny5]

First thing to assume is that y is a simple power function - because each term is in the form x^a*(ath derivative of y). Find all working a's. There should be 3 of them. Let the function be b1*x^a1 + b2*x^a2 + b3*x^a3' date=' where b1, b2, and b3 are any constants. This should solve the equation.

If there are two equal solutions (i.e. a1 = a2), then instead use b1*x^a1+b2*x^a1*ln(x) + b3*x^a3, etc.

-Uncool-[/quote']

 

Thank you very much, I'm still working on it.

 

I remember the ln(x) thing.

 

Can you show me how that goes again?

 

It's in my ODE book, I can find it, but if you happen to know.

 

It's been awhile since I've run through all the various methods of solving ODE's and that was one of the more sophisticated cases.

 

The one where the ln x enters your general solution.

 

What is the 'kind' of ODE that enters?

 

I remember that you need linearly independent solutions, and I remember using the Wronskian to help out. The ln x thing, permits the 'parts' of the solution to be linearly independent, if I remember correctly. It's coming back to me.

 

Thanks

[Johnny5]

First thing to assume is that y is a simple power function - because each term is in the form x^a*(ath derivative of y). Find all working a's. There should be 3 of them. Let the function be b1*x^a1 + b2*x^a2 + b3*x^a3' date=' where b1, b2, and b3 are any constants. This should solve the equation.

If there are two equal solutions (i.e. a1 = a2), then instead use b1*x^a1+b2*x^a1*ln(x) + b3*x^a3, etc.

-Uncool-[/quote']

 

I am going to take this advice and try again to solve the ODE.

 

[math]

3x^3 y^{\prime \prime \prime} + 9x y^\prime - y = 0

[/math]

 

Assume the answer has the following form:

 

[math] y(x) = C_1 x^{a_1} + C_2 x^{a_2} +C_3 x^{a_3} [/math]

 

 

Actually, this is how I started off the problem before, with all but three of the arbitrary constants being equal to zero, so this method is equivalent to the one I used.

 

Hmmm

 

Let me try to solve the following ODE using power series:

 

[math]

 

y^{\prime \prime } + y = 0

 

[/math]

 

Assume that:

 

[math] y(x) = \sum_{n=0}^{n=\infty} C_n x^n [/math]

 

The first derivative is given by:

 

[math] \frac{dy}{dx}= y^{\prime} = \sum_{n=0}^{n=\infty} n C_n x^{n-1} [/math]

 

The second derivative is given by:

 

[math] \frac{d^2y}{dx^2}= y^{\prime \prime} = \sum_{n=0}^{n=\infty} n(n-1) C_n x^{n-2} [/math]

 

Substituting into the orginal expression, we have:

 

[math] \sum_{n=0}^{n=\infty} n(n-1) C_n x^{n-2} + \sum_{n=0}^{n=\infty} C_n x^n = 0 [/math]

 

Performing the necessary indicial operations, we have the following equivalent formula:

 

[math] \sum_{n=-2}^{n=\infty} (n+2)(n+1) C_{n+2} x^n + \sum_{n=0}^{n=\infty} C_n x^n = 0 [/math]

 

And when n=-2, (n+2)=0, and when n=-1, (n+1)=0, so that we may write:

 

[math] \sum_{n=0}^{n=\infty} (n+2)(n+1) C_{n+2} x^n + \sum_{n=0}^{n=\infty} C_n x^n = 0 [/math]

 

Which is equivalent to:

 

[math] \sum_{n=0}^{n=\infty} [(n+2)(n+1) C_{n+2} + C_n ] x^n = 0

[/math]

 

In order for the expression above to denote a true statement when not(x=0), it must be the case that:

 

[math] (n+2)(n+1) C_{n+2} + C_n = 0 [/math]

 

Notice that the gap is 2, thus the equation above will lead to two recfurrence relations. I am going to introduce the variable k for that.

 

One will be even 2k, and the other odd (2k+1).

 

The first thing to do, is to rewrite the formula as follows:

 

[math] (n+2)(n+1) C_{n+2} =- C_n [/math]

 

Now, in the formula above, let n+2=2k, from which it follows that (n+1)=2k-1. Making the substitution yields:

 

[math] (2k)(2k-1) C_{2k} =- C_{2k-2} [/math]

 

When k=1, we have:

 

[math] (2)(1) C_{2} =- C_0 [/math]

 

C0 is one of two arbitrary constants, required to express the general solution to the original ordinary differential equation.

 

Dividing both sides of the recurrence relation by (2k)(2k-1) we have:

 

[math] C_{2k} = \frac{- C_{2k-2}}{(2k)(2k-1) } [/math]

 

Solving the recurrence relation leads to:

 

[math] C_{2n} = C_0 \prod_{k=1}^{k=n} \frac{- 1}{(2k)(2k-1) } [/math]

 

Or equivalently:

 

[math] C_{2n} = (-1)^n C_0 \prod_{k=1}^{k=n} \frac{1}{(2k)(2k-1) } [/math]

 

Now, look at the iterated product formula. What is it as a factorial?

 

The answer comes from looking at what it is doing.

 

I claim the answer is 2n factorial...

 

[math] (2n)! = (2n)(2n-1)(2n-2)....1 [/math]

 

For example, suppose that n=3.

 

In this case 2n=6, so that (2n)! = 6!=6*5*4*3*2*1

 

Now look at the iterated product:

 

[math] \prod_{k=1}^{k=n} \frac{1}{(2k)(2k-1) } [/math]

 

In the case where n=3, we have:

 

[math] \prod_{k=1}^{k=3} \frac{1}{(2k)(2k-1) } = \frac{1}{(2)(1)} \cdot \frac{1}{(4)(3)} \cdot \frac{1}{(6)(5)} = \frac{1}{6!} [/math]

 

Convince yourself that:

 

[math] \frac{1}{(2n)!} = \prod_{k=1}^{k=n} \frac{1}{(2k)(2k-1) } [/math]

 

 

Now, the general solution is going to have the following form:

 

[math] y(x) = \sum_{n=0}^{n=\infty} C_{2n} x^{2n} + \sum_{n=0}^{n=\infty} C_{2n+1} x^{2n+1} [/math]

 

Which is our original assumption since:

 

[math] \sum_{n=0}^{n=\infty} C_{2n} x^{2n} = C_0 + C_2x^2+C_4x^4+...

[/math]

 

AND

 

[math] \sum_{n=0}^{n=\infty} C_{2n+1} x^{2n+1} = C_1x + C_3x^3+C_5x^5+...

[/math]

 

So that:

 

[math] \sum_{n=0}^{n=\infty} C_{2n} x^{2n} + \sum_{n=0}^{n=\infty} C_{2n+1} x^{2n+1} = C_0+C_1x+C_2x^2+C_3x^3+... [/math]

 

Therefore:

 

[math] \sum_{n=0}^{n=\infty} C_{2n} x^{2n} + \sum_{n=0}^{n=\infty} C_{2n+1} x^{2n+1} = \sum_{n=0}^{n=\infty} C_{n} x^{n} [/math]

 

We have already found part of the answer since we found that:

 

[math] C_{2n} = (-1)^n C_0 \prod_{k=1}^{k=n} \frac{1}{(2k)(2k-1) } = C_0 \frac{(-1)^n}{(2n)!} [/math]

 

Thus,

 

[math] y(x) = \sum_{n=0}^{n=\infty} C_0 \frac{(-1)^n}{(2n)!} x^{2n} + \sum_{n=0}^{n=\infty} C_{2n+1} x^{2n+1} [/math]

 

And the first series, is the familiar Taylor series expansion for cos(x).

 

 

Thus:

 

[math] y(x) = C_0 cos(x) + \sum_{n=0}^{n=\infty} C_{2n+1} x^{2n+1} [/math]

 

Now, we just have to do the same thing for the odd coefficients. That is, we need to find the odd coefficients as a function of n, so that we are looking for C _2n+1.

 

For that, we have to return to the original recurrence relation:

 

[math] (n+2)(n+1) C_{n+2} =- C_n [/math]

 

Let n+2=2K+1, from which it follows that n+1=2k.

 

Making the substitutions yield:

 

[math] (2k+1)(2k) C_{2k+1} =- C_{2k-1} [/math]

 

Dividing both sides by (2k+1)(2k) [where we assume not(k=0)] we have:

 

[math] C_{2k+1} = \frac{- C_{2k-1}}{(2k+1)(2k) } [/math]

 

Solving the recurrence relation, we have:

 

[math] C_{2n+1} = C_1 \prod_{k=1}^{k=n} \frac{-1}{(2k+1)(2k) } [/math]

 

Convince yourself that:

 

[math] \prod_{k=1}^{k=n} \frac{1}{(2k+1)(2k) } = \frac{1}{(2n+1)!}[/math]

 

Thus:

 

[math] C_{2n+1} = C_1 \frac{(-1)^n}{(2n+1)!} [/math]

 

From which it follows that we can write our general solution as:

 

[math] y(x) = C_0 cos(x) + \sum_{n=0}^{n=\infty} C_1 \frac{(-1)^n}{(2n+1)!} x^{2n+1} [/math]

 

And the series above, is the sine series. Thus, we can express the general solution as:

 

[math] y(x) = C_0 cos(x) + C_1 sin(x) [/math]

 

And now, we can check our answer.

 

Taking the first derivative, we find that:

 

[math] y^\prime = -C_0 sin(x) + C_1 cos(x) [/math]

 

And taking the second derivative, we find that:

 

[math] y^{\prime \prime} = -C_0 cos(x) - C_1 sin(x) [/math]

 

Now, the original differential equation was:

 

[math] y^{\prime \prime } + y = 0 [/math]

 

And adding using our series solution, we have:

 

[math] y^{\prime \prime} + y = -C_0 cos(x) - C_1 sin(x) + C_0 cos(x) + C_1 sin(x) =0 [/math]

 

So that we have found the general solution to the ODE.

[uncool]

Actually, I meant to assume each separately.

E.g. assume y = x^r first. This will work because it is a linear equation.

In this case:

3*r*(r-1)*(r-2)+9*r-1 = 0

3*r^3 - 9*r^2+15*r-1 = 0

Solve that, and then plug the three r's as a1, a2, and a3 into the equation.

-Uncool-

[Johnny5]

Actually' date=' I meant to assume each separately.

E.g. assume y = x^r first. This will work because it is a linear equation.

In this case:

3*r*(r-1)*(r-2)+9*r-1 = 0

3*r^3 - 9*r^2+15*r-1 = 0

Solve that, and then plug the three r's as a1, a2, and a3 into the equation.

-Uncool-[/quote']

 

Ok

 

[math]

 

3x^3 y^{\prime \prime \prime} + 9x y^\prime - y = 0

 

[/math]

 

Assume that y(x) = x^r

 

[math] y = x^r [/math]

[math] y^\prime = rx^{r-1} [/math]

[math] y^{\prime \prime} = r(r-1)x^{r-2} [/math]

[math] y^{\prime \prime \prime} = r(r-1)(r-2)x^{r-3} [/math]

 

Now plug into the given differential equation.

 

[math] 3x^3 (r(r-1)(r-2)x^{r-3}) + 9x (rx^{r-1}) - x^r = 0

[/math]

 

Which, after using the distributive axiom, becomes:

 

[math] 3 (r(r-1)(r-2)x^{r}) + 9 (rx^{r}) - x^r = 0

[/math]

 

Now, presuming that not(x=0), we can divide both sides by x^r, to obtain:

 

[math] 3r(r-1)(r-2) + 9r - 1 = 0 [/math]

 

Which becomes:

 

[math] 3r(r^2 -3r+2) + 9r - 1 = 0 [/math]

 

Which becomes:

 

[math] 3r^3 -9r^2+6r + 9r - 1 = 0 [/math]

 

[math] 3r^3 -9r^2+15r - 1 = 0 [/math]

 

Hey, this is the formula I got before, only in r, instead of n.

 

Thanks cool.

 

 

Well let me branch then, in the previous approach.

 

I started off assuming a power series solution, in integer powers of x, and reached the following line of work:

 

[math] \sum_{n=0}^{n=\infty}

C_n[3n(n-1)(n-2) + 9n - 1 ]x^n = 0

[/math]

 

Which after some algebra, can be shown to be equivalent to:

 

[math] \sum_{n=0}^{n=\infty}

C_n[3n^3-9n^2+15n-1 ]x^n = 0

[/math]

 

Then, using synthetic division, you can quickly rule out the statement above being true for any integer powers of x. The only possible rational roots to that polynomial in n, are [math] \pm 1, \pm \frac{1}{3} [/math]

 

And so, the only possible positive integer root is 1, which is easily shown to not be a root.

 

Yet, that polynomial is the characteristic equation, for solutions of the form x^r, where r doesn't need to be an integer.