1.5.3.2 Example 2  \(yy^{\prime }=x+y\)

Given

\[ yy^{\prime }=x+y \]

Comparing this to \(\left ( y+g\right ) y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\) shows that

\begin{align*} g & =0\\ f_{0} & =x\\ f_{1} & =1\\ f_{2} & =0\\ f_{3} & =0 \end{align*}

Applying the transformation in (2B) gives

\begin{align*} y & =\frac {1}{u}-g\\ y & =\frac {1}{u}\end{align*}

Results in the ode (4)

\begin{equation} u^{\prime }=k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3} \tag {3}\end{equation}

Where, using (5) gives

\begin{align*} k_{0} & =-f_{3}\\ & =0\\ k_{1} & =3gf_{3}-f_{2}\\ & =0\\ k_{2} & =-3g^{2}f_{3}+2gf_{2}-g^{\prime }-f_{1}\\ & =-1\\ k_{3} & =g^{3}f_{3}-g^{2}f_{2}+f_{1}g-f_{0}\\ & =-x \end{align*}

Hence (3) becomes

\begin{align*} u^{\prime } & =k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3}\\ & =-u^{2}-xu^{3}\end{align*}

Which is Abel ode of first kind. (it is also homogeneous class G). Solving this gives

\[ \ln t-c_{1}+\frac {1}{2}\ln \left ( u^{2}x^{2}+xu-1\right ) -\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {1}{5}\left ( 2xu+1\right ) \sqrt {5}\right ) -\ln \left ( xu\right ) =0 \]

But \(u=\frac {1}{y}\) hence the above becomes

\begin{align*} \ln x-c_{1}+\frac {1}{2}\ln \left ( \frac {x^{2}}{y^{2}}+\frac {x}{y}-1\right ) -\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {1}{5}\left ( \frac {2x}{y}+1\right ) \sqrt {5}\right ) -\ln \left ( \frac {x}{y}\right ) & =0\\ \ln x-c_{1}+\frac {1}{2}\ln \left ( \frac {x^{2}+xy-y^{2}}{y^{2}}\right ) -\ln \frac {x}{y}-\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {\sqrt {5}}{5}\frac {2x+y}{y}\right ) & =0\\ \ln x-c_{1}+\ln \left ( \frac {\sqrt {x^{2}+xy-y^{2}}}{y}\right ) -\ln \frac {x}{y}-\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {\sqrt {5}}{5}\frac {2x+y}{y}\right ) & =0\\ \ln x-c_{1}+\ln \left ( \frac {\frac {\sqrt {x^{2}+xy-y^{2}}}{y}}{\frac {x}{y}}\right ) -\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {\sqrt {5}}{5}\frac {2x+y}{y}\right ) & =0\\ \ln x-c_{1}+\ln \left ( \frac {\sqrt {x^{2}+xy-y^{2}}}{x}\right ) -\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {\sqrt {5}}{5}\frac {2x+y}{y}\right ) & =0\\ \ln x-c_{1}+\frac {1}{2}\ln \left ( \frac {x^{2}+xy-y^{2}}{x^{2}}\right ) -\frac {\sqrt {5}}{5}\operatorname {arctanh}\left ( \frac {\sqrt {5}}{5}\frac {2x+y}{y}\right ) & =0 \end{align*}