3.3.20.1 Reduced Riccati with \(n=-2\)

For the special case of \(n=-2\) the solution can be written directly as given by Eqworld ode0106 as

\begin{equation} y=\frac {\lambda }{x}-\frac {x^{2b\lambda }}{\frac {bx}{2b\lambda +1}x^{2b\lambda }+c_{1}} \tag {1}\end{equation}

Where in the above \(\lambda \) is a root of \(b\lambda ^{2}+\lambda +a=0\).

There is another way to solve the above with \(n=-2\). This can be solved using the substitution

\begin{equation} y=\frac {1}{u} \tag {2}\end{equation}

Hence \(y^{\prime }=-\frac {u^{\prime }}{u^{2}}\) and the ode becomes

\begin{align*} -\frac {u^{\prime }}{u^{2}} & =ax^{-2}+b\frac {1}{u^{2}}\\ -u^{\prime } & =a\frac {u^{2}}{x^{2}}+b\\ u^{\prime } & =-a\frac {u^{2}}{x^{2}}-b \end{align*}

Which is first order Homogeneous ode type (see earlier section). But using (1) is much simpler method as solution can be written directly. The following example shows that using (1) and (2) give same solution.

3.3.20.1.1 Example

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

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.