3.9.6.0.1 Example 1 \(y^{\prime \prime }+\frac {2}{x}y^{\prime }+y=\frac {1}{x}\)

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

In the form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=r\left ( x\right ) \) then \(p=\frac {2}{x},q=1,r=\frac {1}{x}\). 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}}\\ & =1 \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} & =\frac {1}{x}\\ \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} & =\frac {1}{x}\\ \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} & =\frac {1}{x}\\ \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} & =\frac {1}{x}\\ \frac {v^{\prime \prime }}{x}-\frac {v^{\prime }}{x^{2}}-\frac {v^{\prime }}{x^{2}}+\frac {2v^{\prime }}{x^{2}}+\frac {v}{x} & =\frac {1}{x}\\ \frac {v^{\prime \prime }}{x}+\frac {v}{x} & =\frac {1}{x}\\ v^{\prime \prime }+v & =1 \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

\[ v=c_{1}\cos x+c_{2}\sin x+1 \]

Hence

\begin{align*} y & =\frac {v}{x}\\ & =c_{1}\frac {\cos x}{x}+c_{2}\frac {\sin x}{x}+\frac {1}{x}\end{align*}