General solution to a differential equation

Question

Is a differential equation still having a general solution even if the differential equation have a singular solution?

for example:

\begin{aligned} \frac{dy}{dx} = x y^{1/2} \end{aligned}

The Solution: \begin{aligned} y= \left(\frac {1}{4}x^2+c \right)^2 \end{aligned}

But also this singular solution (there is not a constant to obtain it from the above, but is a solution) \begin{aligned} y= 0 \end{aligned}

Is this function correct named as a general solution?: \begin{aligned} y= \left(\frac {1}{4}x^2+c \right)^2 \end{aligned}

Answer 1

What you're hoping for is to name the "general solution" with a multi-parameter family, but that is not always possible for non-linear equations. The general solution of an ODE is the set of all solutions, parametrized by constants or not. For linear equations, since the solutions form an affine space (a linear space translated by some function), they can be parametrized by the form $y_p + c_1 y_1 + \dots + c_n y_n$ where $n$ is the order of the linear equation, $y_p$ a particular solution, $y_1, \dots, y_n$ linearly independent solutions of the homogeneous equation and $c_1, \dots, c_n$ the parameters which generate the general solution.

In non-linear equations you are not that lucky all the time. Sometimes solutions must be picked one by one and some other solutions can be "regrouped" in a parameter family (because you integrated somewhere).

Hope that helps,

Answer 2

Your example has uniqueness problems because $x y^{1/2}$ is not locally Lipschitz as a function of $y$. But consider a differential equation $\frac{dy}{dx} = f(x,y)$ where $f(x,y)$ is continuous in a rectangle $(a,b) \times (c,d)$ and Lipschitz in the $y$ variable there. Then for any $x_0 \in (a,b)$, we can use $y(x_0)$ to parametrize all solutions with $y(x_0) \in (c,d)$. That is, for each $z \in (c,d)$, there is a unique solution $Y(x)$ with $Y(x_0) = z$, defined for $x$ in some maximal interval $(x_1, x_2)\subseteq (a,b)$ such that $Y(x) \in (c,d)$ for $x \in (x_1, x_2)$.