In my Differential Equations class, we began to learn about singular solutions. As I understand, a singular solution of a differential equation (DE) is a solution that cannot be achieved by setting the constant C.
This is my understanding: $$ \frac{dy}{dx} = (y-3)^2 $$ We separate the DE: $$ \frac{1}{(y-3)^2}dy = dx $$ Integrate: $$ \int{\frac{1}{(y-3)^2}dy} = \int{dx}$$ $$ \frac{-1}{y-3} = x + C $$ Solve for y: $$ y = 3 - \frac{1}{x+C} $$
In the original DE, $y =3$ is a solution. However, there is no possible value for $C$ that would lead to this solution. As a result, this is a singular solution.
In class, we were given the following DE and asked to identify possible singular solutions: $$ \frac{dy}{dx} = \frac{6y}{x} $$ We separate and integrate: $$ \int \frac 1 y dy = 6 \int \frac 1 x dx $$ $$ \ln|y| = 6\ln|x| + C_1 $$ Solve for $y$: $$ y = e^{C_1}x^6 $$ $e^{C_1}$ is a new numerical constant and becomes $C$: $$ y = Cx^6 $$
My professor said that at first glance of the original DE $\frac{dy}{dx} = \frac{6y}{x}$, $y = 0$ could be a singular solution (he makes singular solutions in the class turn out to be constants). However, he said that since $C$ could be equal to $0$, $y=0$ is not a singular solution.
So my question is this:
If the original constant of integration was $C_1$, and the new constant of integration $C = e^{C_1}$, how is it possible that $C = 0$ when $e^x$ is never equal to $0$? Is $y = 0$ a singular solution?
Can the newly derived constant $C$ take on any value, despite its origin?
In your second displayed equation, you have divided by $(y-3)^2$. Whenever you divide by a variable quantity you have to split into two cases. Either 1. $y-3 \neq 0$ and so we can divide by it, or 2. $y-3=0$ and we have to deal with that case. The second case gives $y=3$ as a solution, and there is no $C$ in this case.