Find the general solution of the differential equation whose…

Find the general solution of the differential equation whose solution is given by $y=c(x-c)^2$.

Step 1: Differentiate and relate the quantities

Differentiating $y=c(x-c)^2$ gives $y'=2c(x-c)$. Squaring, $(y')^2=4c^2(x-c)^2=4c\cdot\big[c(x-c)^2\big]=4cy$, so

\[c=\frac{(y')^2}{4y}.\]

Also, from $y'=2c(x-c)$, $x-c=\dfrac{y'}{2c}=\dfrac{2y}{y'}$, so $c=x-\dfrac{2y}{y'}$.

Step 2: Eliminate $c$

Equating the two expressions for $c$,

\[x-\frac{2y}{y'}=\frac{(y')^2}{4y}.\]

Multiplying through by $4yy'$ and simplifying gives

\[\boxed{\,(y')^3-4xyy'+8y^2=0.}\]

Step 3: Recover the family and check for a singular solution

By construction, every member of $y=c(x-c)^2$ satisfies this cubic-in-$y'$ equation, since $c$ was eliminated from exactly that family. In addition, $y\equiv0$ (so $y'=0$) satisfies $(0)^3-4x(0)(0)+8(0)^2=0$ identically — this is the singular solution, the envelope ($x$-axis) touched by every parabola $y=c(x-c)^2$ at its vertex $(c,0)$.