Solve y=y'x+(a)/(y')

Solve $y=y'x+\dfrac{a}{y'}$.

Step 1: Identify Clairaut form and general solution

$f(p)=a/p$. For constant $c$, $y=cx+a/c$, $y'\equiv c$. \texttt{sympy} confirms \[y-\left(xy'+\frac{a}{y'}\right)\equiv0\] — the line family is genuine.

Step 2: Singular-solution condition

$f'(p)=-a/p^2$, so $x+f'(p)=0$ gives $x-\dfrac{a}{p^2}=0\Rightarrow p^2=\dfrac{a}{x}$. \texttt{sympy.solve} gives the two real roots \[p=\pm\sqrt{\frac{a}{x}}.\]

Step 3: Substitute both branches (checked symbolically)

For $p=\sqrt{a/x}$: \texttt{sympy.simplify} of $y=xp+a/p$ gives $y=2\sqrt{ax}$, and $y^2$ simplifies to exactly $4ax$. For $p=-\sqrt{a/x}$: $y=-2\sqrt{ax}$, and $y^2$ again simplifies to exactly $4ax$. Both branches agree on the single implicit curve \[y^2=4ax.\]

Step 4: Consistency check

Both sign branches of $p$ independently confirm $y^2-4ax\equiv0$, so the source's two-branch verification is correct and not merely coincidental — it is the same parabola traced by $y=+2\sqrt{ax}$ (upper) and $y=-2\sqrt{ax}$ (lower).