1.3.1.1.2 Example \(y^{\prime }=-x^{-2}+2y^{2}\)

\[ y^{\prime }=-x^{-2}+2y^{2}\]

Comparing this to \(y^{\prime }=ax^{n}+by^{2}\) shows that \(a=-1,b=2,n=-2\). We will first solve this using (1). The quadratic equation is

\begin{align*} b\lambda ^{2}+\lambda +a & =0\\ 2\lambda ^{2}+\lambda -1 & =0 \end{align*}

The roots are \(\frac {1}{2},-1\). Let us pick first \(\lambda =-1\). Hence the solution using (1) is

\begin{align*} y & =\frac {\lambda }{x}-\frac {x^{2b\lambda }}{\frac {bx}{2b\lambda +1}x^{2b\lambda }+c_{1}}\\ & =\frac {-1}{x}-\frac {x^{-4}}{\frac {2x}{-4+1}x^{-4}+c_{1}}\\ & =\frac {-1}{x}-\frac {x^{-4}}{\frac {2}{-3}x^{-3}+c_{1}}\\ & =\frac {1+3c_{1}x^{3}}{2x-3x^{4}c_{1}}\\ & =\frac {1+c_{2}x^{3}}{2x-x^{4}c_{2}}\end{align*}

Let us now try \(\lambda =\frac {1}{2}\). The solution becomes

\begin{align*} y & =\frac {\lambda }{x}-\frac {x^{2b\lambda }}{\frac {bx}{2b\lambda +1}x^{2b\lambda }+c_{1}}\\ & =\frac {1}{2x}-\frac {x^{2}}{\frac {2x}{2+1}x^{2}+c_{1}}\\ & =\frac {1}{2x}-\frac {x^{2}}{\frac {2x^{3}}{3}+c_{1}}\\ & =\frac {3c_{1}-4x^{3}}{4x^{4}+6c_{1}x}\end{align*}

Both these solution verified OK. Now we will solve the same using the transformation \(y=\frac {1}{u}.\)This results in the ode \(y^{\prime }=ax^{n}+by^{2}\) becoming

\begin{align*} u^{\prime } & =-a\frac {u^{2}}{x^{2}}-b\\ u^{\prime } & =\frac {u^{2}}{x^{2}}-2 \end{align*}

We see that this transformation made the ode a homogeneous type which can be easily solved now. This only works for \(n=-2\). Solving this ode gives

\[ u=\frac {-x\left ( 2+c_{1}x^{3}\right ) }{-1+c_{1}x^{3}}\]

Hence

\begin{align*} y & =\frac {1}{u}\\ & =\frac {1-c_{1}x^{3}}{2x+c_{1}x^{4}}\end{align*}

Which is the same as first solution above.