3.15.2.4 Example \(xy^{\prime \prime }+y^{\prime }+Ay=0\)

\begin{equation} xy^{\prime \prime }+y^{\prime }+Ay=0 \tag {1}\end{equation}

Where \(A\) is constant. Multiplying by \(x\) gives

\[ x^{2}y^{\prime \prime }+xy^{\prime }+Axy=0 \]

Comparing the above to (C) \(x^{2}y^{\prime \prime }+\left ( 1-2\alpha \right ) xy^{\prime }+\left ( \beta ^{2}\gamma ^{2}x^{2\gamma }-\left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) \right ) y=0\) shows that

\begin{align*} \left ( 1-2\alpha \right ) & =1\\ Ax & =\beta ^{2}\gamma ^{2}x^{2\gamma }\\ \left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) & =0 \end{align*}

Which implies \(\alpha =0,2\gamma =1\) or \(\gamma =\frac {1}{2}\). Therefore \(\beta ^{2}\gamma ^{2}=A\) gives \(\beta ^{2}=4A\) or \(\beta =2\sqrt {A}\). And \(n=0\). Hence the solution (C1) is

\[ y\left ( x\right ) =c_{1}J_{0}\left ( 2\sqrt {A}\sqrt {x}\right ) +c_{2}Y_{0}\left ( 2\sqrt {A}\sqrt {x}\right ) \]

Alternative and longer method is the following (this is kept just for illustration, as the above method is more direct).

Using the transformation

\[ x=v^{2}\]

Hence

\begin{equation} v=\sqrt {x} \tag {2}\end{equation}

and \(\frac {dv}{dx}=\frac {1}{2\sqrt {x}}\). Therefore

\begin{align} \frac {dy}{dx} & =\frac {dy}{dv}\frac {dv}{dx}\nonumber \\ & =\frac {dy}{dv}\frac {1}{2\sqrt {x}}\nonumber \\ & =\frac {dy}{dv}\frac {1}{2v} \tag {3}\end{align}

And

\begin{align*} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \\ & =\frac {d}{dx}\left ( \frac {dy}{dv}\frac {1}{2v}\right ) \end{align*}

But \(\frac {d}{dx}=\frac {d}{dv}\frac {dv}{dx}\). The above becomes

\begin{align*} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dv}\frac {dv}{dx}\left ( \frac {dy}{dv}\frac {1}{2v}\right ) \\ & =\frac {dv}{dx}\frac {d}{dv}\left ( \frac {dy}{dv}\frac {1}{2v}\right ) \end{align*}

But \(\frac {dv}{dx}=\frac {1}{2\sqrt {x}}=\frac {1}{2v}\). Hence the above becomes

\begin{equation} \frac {d^{2}y}{dx^{2}}=\frac {1}{2v}\frac {d}{dv}\left ( \frac {dy}{dv}\frac {1}{2v}\right ) \tag {4}\end{equation}

But

\[ \frac {d}{dv}\left ( \frac {dy}{dv}\frac {1}{2v}\right ) =\frac {1}{2}\left ( \frac {d^{2}y}{dv^{2}}\frac {1}{v}-\frac {dy}{dv}\frac {1}{v^{2}}\right ) \]

Hence (4) becomes

\begin{equation} \frac {d^{2}y}{dx^{2}}=\frac {1}{4v}\left ( \frac {d^{2}y}{dv^{2}}\frac {1}{v}-\frac {dy}{dv}\frac {1}{v^{2}}\right ) \tag {5}\end{equation}

Substituting (3,5) into (1) gives

\[ x\frac {1}{4v}\left ( \frac {d^{2}y}{dv^{2}}\frac {1}{v}-\frac {dy}{dv}\frac {1}{v^{2}}\right ) +\frac {dy}{dv}\frac {1}{2v}+Ay=0 \]

But \(x=v^{2}\). The above becomes

\begin{align*} \frac {v}{4}\left ( y^{\prime \prime }\frac {1}{v}-\frac {dy}{dv}\frac {1}{v^{2}}\right ) +y^{\prime }\frac {1}{2v}+Ay & =0\\ \frac {1}{4}y^{\prime \prime }-\frac {1}{4}y^{\prime }\frac {1}{v}+y^{\prime }\frac {1}{2v}+Ay & =0\\ \frac {1}{4}y^{\prime \prime }+\frac {1}{4}y^{\prime }\frac {1}{v}+Ay & =0\\ y^{\prime \prime }+y^{\prime }\frac {1}{v}+4Ay & =0 \end{align*}

Multiplying through by \(v^{2}\)

\[ v^{2}y^{\prime \prime }+vy^{\prime }+4Av^{2}y=0 \]

The above of the form

\[ v^{2}y^{\prime \prime }+vy^{\prime }+\left ( a^{2}v^{2}-n^{2}\right ) y=0 \]

Where \(n=0\) and \(a^{2}=4A\) which has the standard solution

\[ y\left ( v\right ) =c_{1}J_{n}\left ( av\right ) +c_{2}Y_{n}\left ( av\right ) \]

Where \(J_{n}\left ( v\right ) \) is the Bessel function of first kind and \(Y_{n}\left ( v\right ) \) is Bessel function of second kind. Since \(v=\sqrt {x}\) and \(a=2\sqrt {A}\) then the solution for (1) becomes (using \(n=0\))

\[ y\left ( x\right ) =c_{1}J_{0}\left ( 2\sqrt {A}\sqrt {x}\right ) +c_{2}Y_{0}\left ( 2\sqrt {A}\sqrt {x}\right ) \]

For example, if \(A=\frac {1}{4}\). Then the ode \(xy^{\prime \prime }+y^{\prime }+\frac {1}{4}y=0\) and the solution above becomes

\[ y\left ( x\right ) =c_{1}J_{0}\left ( \sqrt {x}\right ) +c_{2}Y_{0}\left ( \sqrt {x}\right ) \]