4.1.2.3.4 Example 4

\begin{align*} \left ( x-1\right ) ^{2}y^{\prime \prime }+4xy^{\prime }+2y-2x & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0 \end{align*}

First we need to determine if it is exact or not.  Applying the test

\begin{align*} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) & =0\\ 2-\frac {d}{dx}\left ( 4x\right ) +\frac {d^{2}}{dx^{2}}\left ( \left ( x-1\right ) ^{2}\right ) & =0\\ 2-4+\frac {d}{dx}\left ( 2\left ( x-1\right ) \right ) & =0\\ 2-4+2 & =0\\ 0 & =0 \end{align*}

So it exact.  Comparing this ode 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 & =\left ( x-1\right ) ^{2}\\ g & =4xy^{\prime }+2y-2x\\ & =4xp+2y-2x \end{align*}

Therefore (4) becomes

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =\left ( x-1\right ) ^{2}p+\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\\ 4xp+2y-2x & =\left ( \frac {\partial }{\partial x}\left ( \left ( x-1\right ) ^{2}p\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \left ( x-1\right ) ^{2}p\right ) +\psi _{y}\right ) p\\ 4xp+2y-2x & =2p\left ( x-1\right ) +\psi _{x}+\psi _{y}p\\ 4xp+2y-2x & =p\left ( 2\left ( x-1\right ) +\psi _{y}\right ) +\psi _{x}\end{align*}

Comparing terms shows that

\begin{align*} 4x & =2\left ( x-1\right ) +\psi _{y}\\ 2y-2x & =\psi _{x}\end{align*}

Or

\begin{align*} 2x+2 & =\psi _{y}\\ 2y-2x & =\psi _{x}\end{align*}

Integrating the first equation gives \(\psi =2xy+2y+h\left ( x\right ) \). Differentiating this w.r.t. \(x\) gives \(\psi _{x}=2y+h^{\prime }\left ( x\right ) \). comparing this to the second equation above gives \(2y-2x=2y+h^{\prime }\left ( x\right ) \), hence \(h^{\prime }\left ( x\right ) =-2x\). Hence \(h=-x^{2}+c\). Therefore\(\ \psi =2xy+2y-x^{2}+c\).  Eq (1A) becomes

\begin{align*} R & =\left ( x-1\right ) ^{2}p+2xy+2y-x^{2}+c\\ & =\left ( x-1\right ) ^{2}y^{\prime }+2xy+2y-x^{2}+c \end{align*}

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

\[ \left ( x-1\right ) ^{2}y^{\prime }+2xy+2y-x^{2}=c_{2}\]

Which is the reduced ode to solve.