How to determine if this Differential Equation has a singular solution?

Question

I found this Differential Equation $y'^2=16x^2$ and the text says it has a singular solution...

first they make an implicity derivation with respect to $y'$ then they have $2y'=0$. Therefore $y=constant$ and it proves that the D.E. has a singular solution but I can't figure out why it is correct.

Answer 1

This is a separable equation of the form $y' = f(y) g(x)$. If $f(y_0) = 0$, then $y = y_0$ is an equilibrium solution to the differential equation. However, if there is no equilibrium solution passing through the point $y = y_0$, then $y = y_0$ is a singular solution. In this case, we have $f(y)$ is a constant, so as you stated, $f'(y) = 0$. Therefore it must be true that if $y = y_0$ is an equilibrium solution, it solves the equation $16x^2 = 0$ for all $x$, which is impossible. Therefore, it is a singular solution for any value of $y_0 \in \mathbb{R}$.

Answer 2

Singular Solution is the Envelope of a Family of Curves.
For an equation to be Singular Solution, it has to be common in both p-discriminant and c-discriminant and also has to satisfy the Differential Equation.

Here,
Differential Equation is $ f(x, y, y’) = y'^2 - 16x^2 = 0 $ and
General Solution is $ \phi(x, y, c) = (y-c)^2 - 4x^4 = 0 $

For finding p-discriminant, we find $ \frac {\partial f} {\partial y’} = 0 $ which gives $ 2y’ = 0 $
Now using the above, we need to eliminate $ y’ $ from $ f(x, y, y’) $ which gives $ x^2 = 0 $ which is a possible singular solution.

In this case, when we check whether $ x^2 = 0 $ satisfies the above given Differential Equation, we can see that it doesn’t ( $ \because y’ $ for this curve is tending to $\infty $ )

Graphically also, we can see that $ x^2 = 0 $ is not the envelope of the given family of curves. So, it’s not a singular solution.