The gradient \(y'\) of a curve at a point \((x,y)\) satisfies
\[
(y')^2 -xy'+y=0\,.
\tag{\(*\)}
\]
By differentiating \((*)\) with respect to \(x\), show that either
\(y''=0\) or \(2y'=x\,\).
Hence show that the curve is either a straight line of the form
\(y=mx+c\), where \(c=-m^2\), or the parabola \(4y=x^2\).
The gradient \(y'\) of a curve at a point \((x,y)\) satisfies
\[
(x^2-1)(y')^2 -2xyy'+y^2-1=0\,.
\]
Show that the curve is either a straight line, the form of which you should specify, or a circle, the equation of which you should determine.
Solution:
\(\,\) \begin{align*}
&& 0 &= (y')^2 -xy'+y\\
\Rightarrow && 0 &= 2y' y'' -y' - xy'' + y' \\
&&&= 2y'y'' - xy'' \\
&&&= y'' (2y'-x)
\end{align*}
Therefore \(y'' = 0 \Rightarrow y = mx + c\) or \(y' = \frac12 x \Rightarrow x = \frac14x^2 + C\).
Plugging these into the original equation we have \(m^2 - xm+mx+c = 0 \Rightarrow c = -m^2\)
\(\frac14 x^2 - \frac12 x^2 + \frac14x^2 + C = 0 \Rightarrow C = 0\). Therefore \(4y = x^2\)
\begin{align*}
&& 0 &= (x^2-1)(y')^2 -2xyy'+y^2-1 \\
\Rightarrow && 0 &= 2x(y')^2 +(x^2-1)2y'y'' - 2yy' - 2x(y')^2-2xyy''+2yy' \\
&&&= (x^2-1)2y'y'' -2xyy'' \\
&&&= 2y'' ((x^2-1)y'-xy)
\end{align*}
Therefore \(y'' = 0\) so \(y = mx + c\) or
\begin{align*}
&& \frac{\d y}{\d x} &= \frac{xy}{x^2-1} \\
\Rightarrow && \int \frac1y \d y &= \int \frac{x}{x^2-1} \d x \\
\Rightarrow && \ln |y| &= \frac12 \ln |x^2-1| + C \\
\Rightarrow && y^2 &= A(x^2-1)
\end{align*}
Suppose \(y = mx+c\) then we must have \((x^2-1)m^2-2xm(mx+c)+(mx+c)^2 = -m^2+c^2 \Rightarrow c^2 = m^2\)
If \(y^2 = A(x^2-1)\) then \(2yy' = 2xA\) and
\begin{align*}
&& 0 &= \frac{y^2}{A}\left ( \frac{xA}{y} \right)^2 - 2x^2A+A(x^2-1)-1 \\
&&&= x^2A-2x^2A+x^2A-A-1 \\
\Rightarrow && A &= -1
\end{align*}
Therefore \(x^2 + y^2 = 1\)