Finding the general solution of $xy''+y'+xy=0$ using reduction of order.

Question

Use reduction of order to show that the general solution of $$xy''+y'+xy=0$$ can be written as $$y=c_1J_0(x)+c_2J_0(x) \int^x_1 \frac{1}{s[J_0(x)]^2} ds.$$ Show that if $c_2 \neq 0, y \approx c_2 \ln(x)$ as $x \rightarrow 0^+.$

Notes: First we need to transform the original ODE into Bessel's ODE $x^2y''+xy'+x^2y=0$. As $J_0(x)$ (a 0 order Bessel function of the first kind) is a solution to Bessel's ODE we can write it as the second solution as \begin{align*} y_2&=J_0(x)u(x) \\ y'_2&=J_0'(x)u(x)+J_0(x)u'(x) \\ y''_2 &=J_0^{\prime\prime}u+2J_0^\prime u^\prime+J_0u^{\prime\prime} \end{align*} We can be substituted into $x^2y''+xy'+x^2y=0$ to find the value of $u(x)$

From there my main issues are finding $J_0'(x)$ and $J_0''(x)$, as well as finding the definite integral $\int^x_1 \frac{1}{s[J_0(x)]^2} ds$. How does reduction of order give a definite integral for $u(x)$?

Answer 1

Note that you do not need to find $J_0^\prime$ or $J_0^{\prime\prime}$ to find $u$ and the form $\int_0^x\dfrac{1}{sJ_0(s)}\,ds$ is merely a way of writing the indefinite integral explicitly as a function of $x$. It essentially means the same thing as $\int\dfrac{1}{xJ_0}\,dx$.

\begin{eqnarray} y_2&=&J_0u\\ y_2^\prime&=&J_0^\prime u+J_0u^\prime\\ y_2^{\prime\prime}&=&J_0^{\prime\prime}u+2J_0^\prime u^\prime+J_0u^{\prime\prime} \end{eqnarray}

\begin{equation} x^2y_2^{\prime\prime}+xy_2^\prime+x^2y_2=0 \end{equation}

\begin{equation} x^2\left(J_0^{\prime\prime}u+2J_0^\prime u^\prime+J_0u^{\prime\prime}\right)+x\left(J_0^\prime u+J_0u^\prime\right)+x^2u=0 \end{equation} \begin{equation} (x^2J_0^{\prime\prime}+xJ_0^\prime+x^2J_0)u+x^2(2J_0^\prime u^\prime+J_0u^{\prime\prime})+xJ_0u^\prime=0 \end{equation}

But

\begin{equation} x^2J_0^{\prime\prime}+xJ_0^\prime+x^2J_0=0 \end{equation}

So

\begin{equation} x^2(2J_0^\prime u^\prime+J_0u^{\prime\prime})+xJ_0u^\prime=0 \tag{1} \end{equation}

Substitute $v=J_0u^\prime$. Then $v^\prime=J_0^\prime u^\prime+J_0y^{\prime\prime}$ and $u^\prime=\dfrac{V}{J_0}$.

So equation ($1$) can be re-written as

\begin{eqnarray} x^2\left(\dfrac{J_0^\prime}{J_0}v+v^\prime\right)+xV&=&0\\ x^2v^\prime+\left(x+x^2\cdot\dfrac{J_0^\prime}{J_0}\right)v&=&0\\ v^\prime+\left(\dfrac{1}{x}+\dfrac{J_0^\prime}{J_0}\right)v&=&0 \end{eqnarray}

which has an integrating factor $\mu=xJ_0$ giving $u^\prime J_0=\dfrac{1}{xJ_0}$ so that

\begin{equation} u=\int\dfrac{1}{xJ_0^2}\,dx \end{equation}