3.9.6.0.2 Example 2 \(y^{\prime \prime }+\frac {2}{x}y^{\prime }-y=0\)

\begin{align} y^{\prime \prime }+\frac {2}{x}y^{\prime }-y & =0\tag {1}\\ y\left ( -\infty \right ) & =0\nonumber \\ y^{\prime }\left ( -1\right ) & =-e^{-1}\nonumber \end{align}

In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=\frac {2}{x},q=-1,r=0\). Hence (6A) is

\begin{align*} z & =e^{-\int \frac {p}{2}dx}\\ & =e^{-\int \frac {1}{x}dx}\\ & =e^{-\ln x}\\ & =\frac {1}{x}\end{align*}

Now we check if \(q_{1}\) is constant or a constant divided by \(x^{2}\).

\begin{align*} q_{1} & =q-\frac {1}{2}p^{\prime }-\frac {1}{4}p^{2}\\ & =-1-\frac {1}{2}\left ( \frac {2}{x}\right ) ^{\prime }-\frac {1}{4}\left ( \frac {2}{x}\right ) ^{2}\\ & =-1-\left ( -\frac {1}{x^{2}}\right ) -\frac {1}{4}\frac {4}{x^{2}}\\ & =-1-\left ( -\frac {1}{x^{2}}\right ) -\frac {1}{4}\frac {4}{x^{2}}\\ & =-1+\frac {1}{x^{2}}-\frac {1}{x^{2}}\\ & =0 \end{align*}

Since \(q_{1}\) is constant, then we can use the change of the variable \(y=v\left ( x\right ) z\left ( x\right ) \) which is

\[ y=\frac {v}{x}\]

Since \(z=\frac {1}{x}\). Substituting the above into the original ODE (1) gives

\begin{align*} \left ( \frac {v}{x}\right ) ^{\prime \prime }+\left ( \frac {2}{x}\left ( \frac {v}{x}\right ) ^{\prime }\right ) -\frac {v}{x} & =0\\ \left ( \frac {v^{\prime }}{x}-\frac {v}{x^{2}}\right ) ^{\prime }+\frac {2}{x}\left ( \frac {v^{\prime }}{x}-\frac {v}{x^{2}}\right ) -\frac {v}{x} & =0\\ \left ( \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\left ( \frac {v^{\prime }}{x^{2}}-2\frac {v}{x^{3}}\right ) \right ) +\frac {2}{x}\left ( \frac {v^{\prime }}{x}-\frac {v}{x^{2}}\right ) -\frac {v}{x} & =0\\ \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\frac {v^{\prime }}{x^{2}}+2\frac {v}{x^{3}}+\frac {2v^{\prime }}{x^{2}}-\frac {2v}{x^{3}}-\frac {v}{x} & =0\\ \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\frac {v^{\prime }}{x^{2}}+\frac {2v^{\prime }}{x^{2}}-\frac {v}{x} & =0\\ \frac {v^{\prime \prime }}{x}-\frac {v}{x} & =0\\ v^{\prime \prime }-v & =0 \end{align*}

This is constant coefficient ODE which is easily solved. If the ode in \(v\left ( x\right ) \) did not come to be constant coefficient then we made a mistake. The solution is

\[ v=c_{1}e^{-x}+c_{2}e^{x}\]

Hence

\begin{align} y & =\frac {v}{x}\nonumber \\ & =c_{1}\frac {e^{-x}}{x}+c_{2}\frac {e^{x}}{x} \tag {2}\end{align}

Now we need to find \(c_{1},c_{2}\) from initial conditions. From (2),

\begin{equation} y^{\prime }=-c_{1}\frac {e^{-x}}{x}-c_{1}\frac {e^{-x}}{x^{2}}+c_{2}\frac {e^{x}}{x}-c_{2}\frac {e^{x}}{x^{2}} \tag {3}\end{equation}

Whenever we have \(\infty \) in the IC, we will replace it by \(u\). Hence the IC’s are now

\begin{align} y\left ( -u\right ) & =0\tag {4}\\ y^{\prime }\left ( -1\right ) & =-e^{-1}\nonumber \end{align}

Substituting IC into (2,3) gives two equations to solve for \(c_{1},c_{2}\)

\begin{align*} 0 & =-c_{1}\frac {e^{u}}{u}-c_{2}\frac {e^{-u}}{u}\\ -e^{-1} & =c_{1}e^{1}-c_{1}e^{1}-c_{2}e^{-1}-c_{2}e^{-1}=-2c_{2}e^{-1}\end{align*}

Solving the above two equations for \(c_{1},c_{2}\) gives

\begin{align*} c_{1} & =-\frac {e^{-u}}{2e^{u}}\\ c_{2} & =\frac {1}{2}\end{align*}

But

\[ \lim _{u\rightarrow \infty }\left ( -\frac {e^{-u}}{2e^{u}}\right ) =0 \]

Hence

\begin{align*} c_{1} & =0\\ c_{2} & =\frac {1}{2}\end{align*}

And the solution (2) becomes

\[ y=\frac {1}{2}\frac {e^{x}}{x}\]