I'm trying to solve the following inhomogeneous modified bessel equation. $$y^{\prime\prime}+\frac{1}{x}y^{}\prime-\frac{x^2+4}{x^2}y=x^4$$

I know the homogeneous solution for this differential equation is $y_h=c_1I_2(x)+c_2K_2(x)$

Where $I_2$ and $K_2$ are the modified Bessel function of the first and second kind respectively both of order 2.

For a articular solution i'm trying to get an answer using variation of parameters and full knowing that $W[K_\nu,I_\nu]=1/x$.

Next, i know the particular solution has the form: $$y_p=v_1(x)y_1+v_2(x)y_2$$ where $y_1$ and $y_2$ are the solutions of the homogeneous differential equation respectively.

$v_1(x)=-\int\frac{fy_2}{W}$ and $v_2(x)=\int\frac{fy_1}{W}$ where $f=x^4$

The answer to the problem is give and $y_p=-x^2(x^2+12)$

I don't know how the two integrals can be solved and give something so simple in the end, there's something i'm missing.

  • $\begingroup$ I missed saying that f=x^4 i.e the function at the left hand side $\endgroup$ – Mark A. Ruiz Sep 24 '14 at 5:21
  • $\begingroup$ The problem is the inegrand of $v_1$ and $v_2$ and i just realized something, the integrand of $v_1$ is $x^3K_2(x)=\frac{d}{dx}(x^3K_3)$ I have now both integrals. $\endgroup$ – Mark A. Ruiz Sep 24 '14 at 5:32
  • $\begingroup$ what do you mean by generally? $\endgroup$ – Mark A. Ruiz Sep 24 '14 at 5:33
  • $\begingroup$ Bessel function of the first and second kind yes, sorry i'll edit this $\endgroup$ – Mark A. Ruiz Sep 24 '14 at 5:39
  • $\begingroup$ I just solved them analitically, $v_1(x)=-x^3K_3(x)$ and $v_2=x^3I_3(x)$ $\endgroup$ – Mark A. Ruiz Sep 24 '14 at 5:43


The solution for the associated homogeneous ODE is $y_h=c_1I_2(x)+c_2K_2(x)$

The solution for the non-homogeneous ODE can be found on the form $y=y_h+p(x)$ where $p(x)$ is a particular solution of the ODE.

The seach of a particular solution using the variation parameters method is possible but arduous. Before going on this boring way, it is of use to try some simple functions and see if, by luck, one of them is convenient.

The simplest idea is to try a polynomial. Since there is $-y$ on the left side of the ODE and $x^4$ on the right side, we will try a 4th degree polynomial. Since there is $\frac{-4}{x^2}$ on the left side, the polynomial must not include terms which degree is lower than 2. So, let : $$p(x)=ax^4+bx^3+cx^2$$ Binging it back into the ODE leads to : $$p^{\prime\prime}+\frac{1}{x}p^{\prime}-\frac{x^2+4}{x^2}p= -ax^4-bx^3+(12a-c)x^2+5bx=x^4$$ Hence : $a=-1\space;\space b=0\space;\space c=-12$ We see that, "by luck", the polynomial $p(x)=-x^4-12x^2$ is a convenient particular solution. So, the general solution is : $$y=c_1I_2(x)+c_2K_2(x)-x^4-12x^2$$


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