Can you solve this second-order differential equation with given conditions?

[Mervin]

Hi everyone,

Does anyone know this equation and how to solve it:

(x-a)^2 * (x-b)^2 * y'' = c*y

Thanks

 

[epenguin]

I think the solution is

y = L*(x - A)^((a+K)/(a-b)) . (x - B)^(-(b+K)/(a-b))

Where L, K are arbitrary constants.

Please check (which may be not exactly a doddle).

 

[D H]

Whoa there, epenguin. Please don't give out answers. This site exists in part to help people learn. Telling someone an answer straight out is harmful, not helpful. There is no learning that way.

 

[Mervin]

Thank you very much for the solution. I will check it my self. Don't worry.

Thanks once again!

 

[epenguin]

Whoa there, epenguin. Please don't give out answers. This site exists in part to help people learn. Telling someone an answer straight out is harmful, not helpful. There is no learning that way.

1. I didn't tell him how I got it. (I don't know that this comes into a standard variety of problems. Mervin asked 'does anyone know...' and in a month no one seemed to.) Perhaps it is and it would be good hear from someone, perhaps Mervin, about what the significant generalisations are, and whether there are significant applications.

2. For actually checking it he will have to do an efficiently organised calculation. It took me some time to get right, well it seemed right last night - I would not bet my house on it, so presumably that is a not quite trivial excercise that gives some insights. Mervin?

3. In a sense the only way to solve a differential equation or integration is to know the answer, or at least enough of its shape to be able to hammer it till it fits. I admit to subversive questionings about whether therefore it is a valid thing to teach or give so much emphasis on it and integrations, given that it will be used only by real professionals, and for understanding physics you only need those answers, which you can check.
However I will try to limit my subversion at this site, according to what you say.

4. OK Mervyn, if yo are presented with a d.e. like that, should we say in general f(x).y'' = y , how would you go about solving it, with this one as example?

 

[coomast]

Hello Mervin,

The solution provided by epenguin seems not to be the right one. The factor c is not present, which is an indication. However it can be solved by a transformation. If you use the following:

$$\xi=ln\left|\frac{x-a}{x-b}\right|$$
$$y=(x-b)\cdot \eta$$

You can thransform the given equation into one with constant coefficients, which is easily solved. Let's do this step by step. First use the transformation for y, you get as intermediate results:

$$\frac{dy}{dx}=\eta+(x-b)\frac{d\eta}{dx}$$
$$\frac{d^2y}{dx^2}=2\frac{d\eta}{dx}+(x-b)\frac{d^2\eta}{dx^2}$$

The equation becomes:

$$2(x-a)^2(x-b)\frac{d\eta}{dx}+(x-a)^2(x-b)^2\frac{d^2\eta}{dx^2}=c\eta$$

From the other part of the transformation we have:

$$\frac{d\xi}{dx}=\frac{a-b}{(x-a)(x-b)}$$

 

[coomast]

part two of the solution, I get a database error...

$$\frac{d\eta}{dx}=\frac{a-b}{(x-a)(x-b)} \frac{d\eta}{d\xi}$$
$$\frac{d^2\eta}{dx^2}=-\frac{(a-b)(2x-(a+b))}{(x-a)^2(x-b)^2}\frac{d\eta}{d\xi} +\frac{(a-b)^2}{(x-a)^2(x-b)^2} \frac{d^2\eta}{d\xi^2}$$

The equation becomes now:

$$(a-b)^2\frac{d^2\eta}{d\xi^2}-(a-b)^2\frac{d\eta}{d\xi}-c\eta=0$$

This equation is a second order differential equation with constant coefficients which can be solved in a standard way. The final solution is:

$$y(x)=A|x-a|^{\frac{1}{2}(1+\lambda)}|x-b|^{\frac{1}{2}(1-\lambda)} + B|x-a|^{\frac{1}{2}(1-\lambda)}|x-b|^{\frac{1}{2}(1+\lambda)}$$

with:

$$\lambda=\sqrt{1+\frac{4c}{(a-b)^2}}$$

After all these weeks I finally found it in one of my recently obtained books. The book "Handbook of Exact Solutions for Ordinary Differential Equations" written by Polyanin has this one described. The method is given, but not how the transformation was obtained. However I suspect that it is closely related to Lie point symmetries, which is a theory (new for me) I am starting to learn. I can't use it at this equation yet, but maybe in the future I will be able to show that it is the correct way. Can you give some more information on where this equation is used or coming from?

 

[epenguin]

Well I got the hard part right!



I had another look at this equation

(x-a)^2 * (x-b)^2 * y'' = c*y. (1)

For the solution, obtained without a book by a different argument than coomast’s, I gave

y = L*(x - a)^((a+K)/(a-b)) . (x - b)^(-(b+K)/(a-b)) (2)

It is true of this, as can be checked by twice differentiating, that

(x-a)^2 * (x-b)^2 * y'' = C*y (3)

where you can get any C you want by choosing the K. So, late at night I thought therefore 'I've done it'. Not quite right or complete. (It was subliminally disquieting that my formulation after that was so unlike the linear constant coefficient case.)

The way to think of it rather is, if you twice differentiate w.r.t. x, the function that appears in eq. 2 above without the constant, i.e. define

g(x, K) = (x - a)^((a+K)/(a-b)) . (x - b)^(-(b+K)/(a-b)) (4)

you find

(x-a)^2 * (x-b)^2 *g’’(x, K) = (a + K)*(b + K)*g(x, K) (5)

(More explicitly

(x-a)^2 * (x-b)^2 * [(x - a)^((a+K)/(a-b)) . (x - b)^(-(b+K)/(a-b))]’’

= (a + K)*(b + K)*[(x - a)^((a+K)/(a-b)) . (x - b)^(-(b+K)/(a-b))] ) (6)

So this function g is a solution of

(x-a)^2 * (x-b)^2 * y'' = C*y

if (a + K)*(b + K) = C (7)

There are then two values of K, K1, K2 say, that satisfy (7). So

y = g(x, K1) and y = g(x, K2) are solutions and so is

y = A*g(x, K1) + B*g(x, K2) (8)

where A, B are arbitrary constants in the usual way remembered from lde’s.

When we incorporate the K’s which are solutions of (7), the formula is exactly the same as that given by coomast.

As this has turned out not all that easy maybe it is OK to give my way of finding the solution and rewrite in LaTex presently?

 

[epenguin]

Hello. I thought after all this time and as a solution, I think identical to mine has been posted, I would indicate how I got it. The idea is basically that the form eq. (2) below is a plausible one natural to try, even if with a bit more insight (or hindsight) another form leads you there too.

I have had trouble with latex posting (database error and I have to trick the system by building up a bit at a time) so I now have to do this in two posts. Hopefully, cough cough, here goes:


To

$$(x - a)^2(x-b)^2y^\prime^\prime = cy\ (1)$$

assume a solution of form

$$y = e^p$$. (2)


(to be continued...)

 

[epenguin]

Then

$$y^\prime^\prime = (e^p)^\prime^\prime$$

$$\ = (p^\prime^\prime + p\prime^2)e^p$$

Combining with (1) we obtain

$$p^\prime^\prime + p^\prime^2 = \frac{c}{(x - a)^2(x-b)^2}$$ (3)

In order to get such a demominator it seems necessary or probable we need a denominator (x - a)(x-b) in $$p^\prime$$. So let

$$p^\prime = \frac{f(x)}{(x - a)(x-b)}$$ (4)

Then

$$p^\prime^\prime + p^\prime^2 = \frac{(x-a)(x-b)f^\prime - [(x-a) + (x-b)]f + f^2}{(x - a)^2(x-b)^2}$$ (5)

Assume f is a polynomial of degree n. Then the first two numerator terms are of degree (n + 1) whilst the last is of degree 2n. For a cancellation to leave only a constant as in eq.(3) to be possible, it is necessary that n=1. So let

$$f = lx + K$$ (6)

The numerator is then found to be

$$x^2(l^2-l) + 2xK(l-1) + (abl + aK + bK + K^2)$$

so to give form (3)

$$l = 1$$ and

$$ab + (a+b)K +K^2 = c$$ (7)

and (4) becomes

$$p^\prime = \frac{x+K}{(x - a)(x-b)}$$.

So

$$p = \int\frac{x+K}{(x - a)(x-b)}\ dx$$

$$= \frac{(a+K)\ln(x-a) - (b+K)\ln (x-b)}{(a-b)} + K^\prime$$

So $$y = e^p$$ works out as

$$y = (x-a)^\frac{a+K}{a-b} (x-b)^-^\frac{b+K}{a-b}$$

This y multiplied by any constant is also a solution, and is so for either of the values of K which are solutions of eq. (7), so similarly to the way of solving lde's, and inserting the solutions for K, we find the general solution

$$y = A(x-a)^r(x-b)^s + B(x-a)^s(x-b)^r$$

where $$r,s = \frac{1}{2}[1 \pm\surd(1+\frac{4c}{(a-b)})]$$

and A, B arbitrary constants.