3.3.21.3 Examples
3.3.21.3.1 Example 1

3.3.21.3.1 Example 1

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

Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) form shows that \(f_{0}=-x,f_{1}=0,f_{2}=\frac {1}{x}\). We will use the method of converting to second order ode. Let \(y=\frac {-u^{\prime }}{f_{2}u}=x\frac {u^{\prime }}{u}\). Using this substitution results in

\begin{align*} f_{2}u^{\prime \prime }-\left ( f_{2}^{\prime }+f_{1}f_{2}\right ) u^{\prime }+f_{2}^{2}f_{0}u & =0\\ \frac {1}{x}u^{\prime \prime }-\left ( -\frac {1}{x^{2}}\right ) u^{\prime }+\left ( \frac {1}{x^{2}}\right ) \left ( -x\right ) u & =0\\ \frac {1}{x}u^{\prime \prime }+\frac {1}{x^{2}}u^{\prime }-\frac {1}{x}u & =0\\ xu^{\prime \prime }+u^{\prime }-xu & =0 \end{align*}

This is Bessel ode the solution is

\[ u=c_{1}\operatorname {BesselI}\left ( 0,x\right ) +c_{2}\operatorname {BesselK}\left ( 0,x\right ) \]

But \(y=x\frac {u^{\prime }}{u}\), hence

\[ y=x\frac {\left ( c_{1}\operatorname {BesselI}\left ( 1,x\right ) -c_{2}\operatorname {BesselK}\left ( 1,x\right ) \right ) }{c_{1}\operatorname {BesselI}\left ( 0,x\right ) +c_{2}\operatorname {BesselK}\left ( 0,x\right ) }\]