Singular solution of differential equation $p^2y+2px-y=0$

Question

I have the following differential equation before me for which I am required to find singular solution:

$p^2y+2px-y=0$ where $p=\dfrac{dy}{dx}$

On solving it, I got the following general solution:

$y^2=c^2+2cx$

We get $p$-discriminant and $c$-discriminant relation( they turn out to be same in this case) as:

$x^2+y^2=0$

This is a point circle centered at origin. Can it be regarded as singular solution? Our general solution is a family of parabolas and $x^2+y^2=0$ does not touch any of them. Does it mean that for this differential equation, there is no singular solution.

Please help.

Answer 1

Use a solution with a general approach to compare your solution against.

Multiply the equation with $y$ to get $$ y^2 = (2py)x+\frac14(2py)^2. $$ This is a Clairaut equation for $v=y^2$, $$ v=v'x+\frac14(v')^2. $$ The singular solution in $v$ is given via the equation $0=x+\frac12v'$, thus $v'=-2x$ and directly inserted $$ v=-x^2. $$ As the part of that with non-negative $v$ is just a point, there is no way to observe the tangency of any curve to it when resolved to the original $y$.

So indeed all you calculated and observed is true.