4.1.2.3.1 Example 1

\[ yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2}=0 \]

Comparing this to  \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =y\\ g & =\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2}\\ & =p^{2}+2axyp+ay^{2}\end{align*}

Therefore (4) becomes

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =yp+\psi \left ( x,y\right ) \tag {1A}\end{align}

Hence (5) becomes

\begin{align*} g & =\left ( \frac {\partial }{\partial x}\left ( \int fdp\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \int fdp\right ) +\psi _{y}\right ) p\\ p^{2}+2axyp+ay^{2} & =\left ( \frac {\partial }{\partial x}\left ( yp\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( yp\right ) +\psi _{y}\right ) p \end{align*}

But \(\frac {\partial }{\partial x}\left ( yp\right ) =0\) since \(y,p\) are held constant. It is important to watch for this here. Given \(f\left ( x,y\right ) =3x+y\left ( x\right ) \) where \(y\) is function of \(x\), then when we do \(\frac {\partial f}{\partial x}\) the result is \(3\) and not \(3+y^{\prime }\) because with partial derivatives the \(y\) is held constant. Similarly \(\frac {\partial }{\partial y}\left ( yp\right ) =p^{2}\). The above becomes

\begin{align*} p^{2}+2axyp+ay^{2} & =\psi _{x}+\left ( p+\psi _{y}\right ) p\\ & =\psi _{x}+p^{2}+\psi _{y}p\\ 2axyp+ay^{2} & =\psi _{x}+\psi _{y}p \end{align*}

Comparing terms shows that

\begin{align} 2axy & =\psi _{y}\tag {2A}\\ ay^{2} & =\psi _{x} \tag {3A}\end{align}

Integrating (2A) w.r.t \(y\) gives

\begin{equation} \psi =axy^{2}+h\left ( x\right ) \tag {4A}\end{equation}

Differentiating the above w.r.t. \(x\) gives \(\psi _{x}=ay^{2}+h^{\prime }\left ( x\right ) \). comparing this to (3A) above gives \(ay^{2}=ay^{2}+h^{\prime }\left ( x\right ) \), hence \(h^{\prime }\left ( x\right ) =0\) or \(h\left ( x\right ) =c\). Therefore (4A) becomes

\[ \psi =axy^{2}+c \]

Substituting the above in (1A) gives

\[ R=yp+axy^{2}+c \]

Therefore, since \(R=c_{1}\) a constant, then the above becomes (by merging the constants)

\begin{align*} yp+axy^{2} & =c_{2}\\ yy^{\prime }+axy^{2} & =c_{2}\end{align*}

This is the reduced ode which needs to be solved for \(y\). The above says that \(R=yy^{\prime }+axy^{2}+c_{2}\). To verify, let us apply \(F=\frac {d}{dx}R\). This gives

\begin{align*} yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2} & =\frac {d}{dx}\left ( yy^{\prime }+axy^{2}+c_{2}\right ) \\ & =y^{\prime }y^{\prime }+yy^{\prime \prime }+ay^{2}+2axyy^{\prime }\\ & =yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+2axyy^{\prime }+ay^{2}\end{align*}

Verified.