3.3.21.3 Examples

3.3.21.3.1 Example 1

$$\begin{array}{}\text{(1)}& y^{\prime }=-x+\frac{1}{x}{y}^2\end{array}$$

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

$$\begin{array}{rl}{f}_2{u}^{\prime \prime }-(f_2^{\prime }+f_1{f}_2)u^{\prime }+f_2^2{f}_{0}u& =0\\ \frac{1}{x}{u}^{\prime \prime }-(-\frac{1}{x^2})u^{\prime }+\left(\frac{1}{x^2}\right)(-x)u& =0\\ \frac{1}{x}{u}^{\prime \prime }+\frac{1}{x^2}{u}^{\prime }-\frac{1}{x}u& =0\\ x{u}^{\prime \prime }+u^{\prime }-xu& =0\end{array}$$

This is Bessel ode the solution is

$$u=c_1\mathrm{BesselI}(0,x)+c_2\mathrm{BesselK}(0,x)$$

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

$$y=x\frac{(c_1\mathrm{BesselI}(1,x)-c_2\mathrm{BesselK}(1,x))}{c_1\mathrm{BesselI}(0,x)+c_2\mathrm{BesselK}(0,x)}$$