Is this a famous differential equation?

Question

Hi guys!

I'm looking at the following equation:

y''+Cxy'+Dx^(2)y=0

where ' denotes differentiation with respect to x and C and D are positive constants. What are the solutions to this equation? Is it named after an old man or woman?

Thanks!

Thanks for the varied and interesting responses!

I'm surprised by the specific result from Mathematica Mistakes/ . I managed to install a solver on my computer and it would appear that the kinds of functions you describe are an envelope to cos^(2)(x^(2)) type oscillation - i.e. the solution appears as a rapidly damped oscillator of ever decreasing period. Indeed, as C*x and D*x^(2) are inherently positive, is the system rather like an ever more damped harmonic oscillator?

For the curious, the equations describe the response of a certain field in a cosmological background to small perturbations!

Answer 1

This is a very interesting linear equation. As noted, the equation x^2 y''+cx y'+dy=0 is called a Caucy-Euler equation.

I am not sure if this has a name or not.

Mathematica could not find a general solution for arbitrary c and d. However, if you give it particular values, in some cases it can. For example, if C=2 and D=1 it returns

y=1/2 exp(-x-x^2/2)(2C[1]+exp(2x) C[2])

as a general solution. (C[1] and C[2]) represent the arbitrary constant.

Assumiing that y=exp(f(x)), substituting into the equation results in

exp(f(x))(dx^2+cxf'(x)+f'(x)^2+f''(x))=0

so

dx^2+cxf'(x)+f'(x)^2+f''(x)=0

To get this to work, f(x) is almost certainly a polynomial (because we will have to kill off the dx^2 term). Moreover, the highest degree it could be is 2 because then cxf'(x) and f'(x)^2 would be of degree 2 (to kill of the dx^2 term).

Now, let's assume that f(x)=a_2 x^2 +a_1 x. (We don't need a constant because the derivative of the constant will be 0.) Substiuting into the above gives us

(4a_2^2+2c a_2+d)x^2+(ca_1+4a_1a_2)x+a_1^2+2a_2=0

When we can solve
4a_2^2+2c a_2+d=0
ca_1+4a_1a_2=0
+a_1^2+2a_2=0

we will have a solution of the form exp(a_2 x^2+a_1x)

That is cool! :)

I cannot spot an obvious mistake... But, I will think about this more... Very interesting...

Here is more: Notice that if we let u=f'(x),

dx^2+cxf'(x)+f'(x)^2+f''(x)=0

becomes the first order equation

u'+cxu+u^2=-dx^2

Here is Mathematica's solution

In[6]:=
DSolve[u'[x]+cx u[x]+u[x]^2\[Equal]-d x^2,u[x],x]//Simplify//InputForm
Out[6]//InputForm=
{{u[x] -> ((-1)^(3/4)*(cx^2 + (4*I)*Sqrt[d])*C[1]*
HermiteH[-3/2 + ((I/8)*cx^2)/Sqrt[d],
(-1)^(1/4)*d^(1/4)*x] +
d^(1/4)*(-2*(cx + (2*I)*Sqrt[d]*x)*C[1]*
HermiteH[-1/2 + ((I/8)*cx^2)/Sqrt[d],
(-1)^(1/4)*d^(1/4)*x] - 2*(cx + (2*I)*Sqrt[d]*x)*
Hypergeometric1F1[1/4 - ((I/16)*cx^2)/Sqrt[d],
1/2, I*Sqrt[d]*x^2] + (cx^2 + (4*I)*Sqrt[d])*x*
Hypergeometric1F1[5/4 - ((I/16)*cx^2)/Sqrt[d],
3/2, I*Sqrt[d]*x^2]))/(4*d^(1/4)*
(C[1]*HermiteH[-1/2 + ((I/8)*cx^2)/Sqrt[d],
(-1)^(1/4)*d^(1/4)*x] + Hypergeometric1F1[
1/4 - ((I/16)*cx^2)/Sqrt[d], 1/2,
I*Sqrt[d]*x^2]))}}

So, one could recover f by integrating the above.

Thanks for saying where you got it at.

I would be interested in working on this equation more with you, if you like. Perhaps, it would result in an interesting publication? My physics is rusty. But, my differential equations and Mathematica skills are reasonable. :)

I have plotted the solution for c and d values 1 to 10 (100 total combinations) and initial conditions -5 to 5 in increments of 2 (36 total) for a total of 3600 solutions. _All_ of them go to 0. Is there a painless (or even obvious) way to prove that y=0 is globally stable for this equation?

Answer 2

The general solution in terms of a series is as follows:
y = u*exp(ax^2) where u is given by
u = c_1[1 + sum from m=1 to m=inf {(-1)^m*(q)(2p+q)...[2p(m-1)+q]x^2m}/(2m)!
+c_2[x + sum from m=1 to m=inf{(-1)^m*(p+q)(3p+q)...[p(2m-1)+q]x^(2m+1)}/(2m+1)!
where p=sqrt(C^2 - 4D), a = (p - C)/4, and q = 2a.
This solution assumes that C^2 - 4D is non-negative. If it is negative then, probably, the exponential is replaced by sine and cosine of ax^2. I have not checked that. I note that the website does not write out all the terms in the sums that I have typed. I hope that you can fill in any missing terms. I now realize that u is a Hermite function with a suitable redefinition of the variables.

Answer 3

If you factor out an x term to get

y'' + x*(Cy'+Dxy) = 0 it's in the form

y'' + x*f(x,y,y') = 0

which looks 'kinda' familiar but I can't place it. It may well be named after somebody. But it isn't Gauss, Legendre, Bessel, or any of the 'big name' superstars.

Maybe one of the hypergeometric functions?


Doug

Answer 4

I don't know about this form but there is another one, which is opposite of this form

x^2y"+xy'+y=0

which is called the Cauchy-Euler equation (and yes, they are both old guys) and has a solution in the form of y=x^r.

Answer 5

y''' + C*x*y''*y' + 2*D*x*y*y' = 0

Answer 6

chap