# Inhomogeneous modified Bessel differential equation

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.

• I missed saying that f=x^4 i.e the function at the left hand side – Mark A. Ruiz Sep 24 '14 at 5:21
• 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. – Mark A. Ruiz Sep 24 '14 at 5:32
• what do you mean by generally? – Mark A. Ruiz Sep 24 '14 at 5:33
• Bessel function of the first and second kind yes, sorry i'll edit this – Mark A. Ruiz Sep 24 '14 at 5:39
• I just solved them analitically, $v_1(x)=-x^3K_3(x)$ and $v_2=x^3I_3(x)$ – Mark A. Ruiz Sep 24 '14 at 5:43

$$y^{\prime\prime}+\frac{1}{x}y^{\prime}-\frac{x^2+4}{x^2}y=x^4$$
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 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 : $$y=c_1I_2(x)+c_2K_2(x)-x^4-12x^2$$