3.1.1.4 Example 4

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

\(f\left ( x,y\right ) =\frac {-1}{1+x}y^{2}+\frac {1}{x-1}\) is continuous in \(x\) everywhere except at \(x=-1\) and \(x=1\). And \(f_{y}=\frac {-2}{1+x}y\) is continuous except at \(x=-1\). Since initial conditions at \(x_{0}=0,y_{0}=0\) then there is a unique solution in some rectangle inside the rectangle \(-1<x<1\) and for all \(y\). Solving the ode gives

\[ 2\sqrt {y}=\int _{0}^{x}\frac {\sqrt {y\sin \tau }}{\sqrt {y}}+c_{1}\]

At \(x=0,y=0\) the above gives

\[ 0=c_{1}\]

Hence the solution is

\[ 2\sqrt {y}=\int _{0}^{x}\frac {\sqrt {y\sin \tau }}{\sqrt {y}}\]