2
$\begingroup$

How can I find a particular solution to this equation?

$$f''(x)+x^{-1}f'(x)-f(x)\left(1+x{}^{-2}\right)-Ax^{-2}=0$$

I know that the solution of the homogeneous equation (i.e. without the last term) is of the form

$$f(x)=c_{1}K_{1}(x)+d_{1}I_{1}(x)$$

but can't work out how to construct a particular solution. Variation of parameters? Thanks!

$\endgroup$
14
  • $\begingroup$ Variation of parameters should work. You might try some identity involving Bessel functions, orthogonality and $x^{-2}$. $\endgroup$
    – Pragabhava
    Commented Jan 11, 2016 at 15:30
  • $\begingroup$ @Pragabhava Could you expand on that please? $\endgroup$ Commented Jan 11, 2016 at 15:33
  • 1
    $\begingroup$ @Pragabhava They can be imposed at the end (at least modulo singularities). $\endgroup$
    – Ian
    Commented Jan 11, 2016 at 15:37
  • 1
    $\begingroup$ OK, but I have to warn you that in my experience as student and teacher, it's a very undidactic way to learn (and work with) odes and pdes. A differential equation is not properly defined without boundary conditions and a domain of integration. $\endgroup$
    – Pragabhava
    Commented Jan 11, 2016 at 15:45
  • 1
    $\begingroup$ @Pragabhava A linear ODE on a fixed domain can be usefully defined without requiring boundary conditions. The solutions form an affine space, which is the reason we seek particular solutions and homogeneous solutions separately when we attempt to identify general solutions. When it is possible to identify general solutions, this is a good technique for solving IVPs/BVPs, since adjoining the initial or boundary conditions then amounts to solving a system of linear equations. Specific boundary conditions are more important in PDE for many reasons, but the problem at hand here is an ODE. $\endgroup$
    – Ian
    Commented Jan 11, 2016 at 16:24

1 Answer 1

1
$\begingroup$

Let's say that you have initial conditions at $x = a$ and your domain of integration is $x\in(a,b)$. The variation of parameters formula says that: $$ f_p(x) = - A I_1(x)\int_a^x \frac{K_1(t) dt}{W(t) t^2} + A K_1(x)\int_a^x \frac{I_1(t)dt}{W(t)t^2}, $$ where $W(t)$ is thhe Wronskian of $K_1(t)$ and $I_1(t)$, which is1 $$ W(t) = K_1(t)I_1'(t) - K_1'(t) I_1(t) = \frac{1}{t}. $$

And so $$ f_p(x) = -A K_1(x)\int_a^x \frac{I_1(t) dt}{t} + A I_1(x)\int_a^x \frac{K_1(t)dt}{t}. $$

which can, in turn, be explicitly written in terms of the Meijer G and the Generalized hypergeometric function.

$\endgroup$
4
  • $\begingroup$ You wrote the coefficients of the homogeneous solutions in the particular solution but you neglected to multiply them by the homogeneous solutions themselves. $\endgroup$
    – Ian
    Commented Jan 11, 2016 at 16:35
  • $\begingroup$ @Ian True. I've edited accordingly. $\endgroup$
    – Pragabhava
    Commented Jan 11, 2016 at 16:50
  • $\begingroup$ The minus sign is in front of the wrong term, I guess. $\endgroup$ Commented Jan 11, 2016 at 17:18
  • 1
    $\begingroup$ @usumdelphini Sorry about that. It has been corrected. $\endgroup$
    – Pragabhava
    Commented Jan 11, 2016 at 17:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .