4.4.2.4 Example 4

Solve

\begin{align} yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}-y^{\prime } & =0\tag {1A}\\ yy^{\prime \prime }+y^{\prime }\left ( y^{\prime }-1\right ) & =0 \tag {1B}\end{align}

This ode can also be solved using the method of missing \(x\). Comparing the above to (1B)

\begin{equation} a_{2}\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+a_{1}\left ( x,y,y^{\prime }\right ) y^{\prime }+a_{0}\left ( x,y,y^{\prime }\right ) =0 \tag {1}\end{equation}

shows that

\begin{align} a_{2} & =y\tag {2A}\\ a_{1} & =\left ( y^{\prime }-1\right ) \nonumber \\ a_{0} & =0\nonumber \end{align}

Note that there is ambiguity in this method in terms of what to use for \(a_{0},a_{1}\). It is possible to read the above ode as having the following pattern. Looking at (1A) now, then

\begin{align} a_{2} & =y\tag {2B}\\ a_{1} & =y^{\prime }\nonumber \\ a_{0} & =-y^{\prime }\nonumber \end{align}

It is also possible to use this third matching

\begin{align} a_{2} & =y\tag {2C}\\ a_{1} & =0\nonumber \\ a_{0} & =\left ( y^{\prime }\right ) ^{2}-y^{\prime }\nonumber \end{align}

These three are all valid matches. How to know which one to use assuming they are verify the exactness conditions? Pick the one that satisfy the exactness conditions. If there is more than one that satisfy the exactness conditions, any one will do.  Let us try the last match (2C) above for now and see. We first verify the ode is exact using the conditions

\begin{align*} \frac {\partial a_{2}}{\partial y} & =\frac {\partial a_{1}}{\partial y^{\prime }}\\ \frac {\partial a_{2}}{\partial x} & =\frac {\partial a_{0}}{\partial y^{\prime }}\\ \frac {\partial a_{1}}{\partial x} & =\frac {\partial a_{0}}{\partial y}\end{align*}

This gives

\[ 1=0 \]

So match (2C) did not work. Lets now use the first match above (2A). We first verify the ode is exact using the conditions. This now gives

\begin{align*} 1 & =1\\ 0 & =0\\ 0 & =0 \end{align*}

So match (2A) verified the exactness. Using this and since no initial conditions are given, then we will use (4). This gives

\begin{align*} \int _{0}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{0}^{y}a_{1}\left ( 0,\beta ,y^{\prime }\right ) d\beta +\int _{0}^{y^{\prime }}a_{2}\left ( 0,0,\gamma \right ) d\gamma & =c_{1}\\ 0+\left ( y^{\prime }-1\right ) \int _{0}^{y}d\beta +\int _{0}^{y^{\prime }}\left ( 0\right ) d\gamma & =c_{1}\\ \left ( y^{\prime }-1\right ) y & =c_{1}\\ y^{\prime }y-y & =c_{1}\\ y^{\prime } & =\frac {c_{1}+y}{y}\end{align*}

Integrating

\begin{align*} \frac {dy}{\frac {c_{1}+y}{y}} & =dx\\ \int \frac {y}{c_{1}+y}dy & =\int dx\\ y-c_{1}\ln \left ( y+c_{1}\right ) & =x+c_{2}\end{align*}

Which is the correct solution to the original ode.

Lets us now try match (2B). First we need to verify it satisfies the exactness conditions.

\begin{align*} \frac {\partial a_{2}}{\partial y} & =\frac {\partial a_{1}}{\partial y^{\prime }}\\ \frac {\partial a_{2}}{\partial x} & =\frac {\partial a_{0}}{\partial y^{\prime }}\\ \frac {\partial a_{1}}{\partial x} & =\frac {\partial a_{0}}{\partial y}\end{align*}

If we now try match (2B) above, which is \(a_{2}=y,a_{1}=y^{\prime },a_{0}=-y^{\prime }\) then the above gives

\begin{align*} 1 & =1\\ 0 & =-1\\ 0 & =0 \end{align*}

Hence this match does not satisfy the exactness conditions. So out of the three possible matches (2A,2B,2C) only (2A) can be used and this gives the correct solution.