# General solution to a 2nd-order differential equation

• I
Dear all

I've been trying to work out the general solution to a 2nd order ODE of the form
f''(x)+p(x)*f(x)=0
p(x) is a polynomial for my case. I believe series method should work, but for some reason I would prefer a general solution using other methods. I'll be much appreciative for any help.

sincerely,
Yo-Ming Cheng

Buzz Bloom
Gold Member
Hi piggees:

Here is an approach that may help you. I am a little rusty on notation, but you can probably find it readable.
D is an differential operator:
D = d/dx
D[f] = df/dx
D2[f] = D×D[f] = d2f/dx

Your equation can be written as
(D2 +p)[f] = 0

I is the identity operator
R is the square root operator
R2 = I

(D2 +p×I) [f] = f" + pf

(D2 +p×I) = (D2 +√p×R) × (D2 -√p×R)

(D2 +p×I) [f] = (D +√p×R)[f] × (D - √p×R)[f]

Therefore one solution of f''+pf=0 is the solution of (D +√p×R)[f] = 0
and the other solution is the solution of (D - √p×R)[f] = 0.

Hope this helps.

Regards,
Buzz

• piggees

Buzz Bloom
Gold Member
Hi piggees:

I want to apologize for the errors in my post. It has been quite a while since I worked with differential equations, and I seem to have misremembered quite a bit.
First my factorization was wrong since I omitted i=sqrt(-1).
Second, it is wrong to take the square root of f in the factorization.
Third, The method of factorization in general has limitations. For example, if the coefficients are constants it is OK, and maybe also under some other special conditions.

Fixing the first and second errors leads to omitting the I and R operators and adding the needed i occurrences.

(D2 +p) [f] = f" + pf

(D2 +p) = (D + i√p) × (D2 -i√p)

(D2 +p) [f] = (D + i√p)[f] × (D - i√p)[f]

However, this factorization only works for p = constant, and possibly also some other very special cases which I no longer remember how to identify.

I would be curious to see in what way you found all my mistakes useful. I apologize again for wasting your time.

Regards,
Buzz

• jim mcnamara
Hi, Buzz, just no need to apologize. I perceived the mistaken factorization in your original reply right away, but I got your point and fixed it myself. I intended to achieve a solution (but not series solution) for a general polynomial p, yet by practicing your method I thought that almost impossible so now I restrict p to 2nd order. This makes general solution to (D + i√p)[f]=0 and (D - i√p)[f]=0 possible. If I'm so lucky that you may provide knowledge as to the solution with a general p, must let me know. I appreciate your suggestion.

Best wishes,
Yo-Ming Cheng

I want to know about the solution of 3rd order ordinary differential equation with variable coeffient
x^3Y''' +5x^2Y'' +7xY'+8Y=0

bobob
Gold Member
I want to know about the solution of 3rd order ordinary differential equation with variable coeffient
x^3Y''' +5x^2Y'' +7xY'+8Y=0
As my math methods instructor use to say, use the staring method. First you stare at it and if nothing comes to mind, stare at it some more. Apart from doing that and seeing something that makes the equation simple to solve, I would just start with a series solution and see if anything about the series lets me write a solution in closed form. In this case, the powers of x are the same in every term, so it looks like you have to solve a cubic:

a_n x^n [ n (n-1)(n-2) + 5 n (n-1) + 7n + 8 ] = 0

Svein