1.4 Existence

Intuitively, if $f(x,y)=\frac{dy}{dx}$, and $f$ is continuous on a bounded rectangle $R:a<x<b,c<y<d$ containing points $(x_0,y_0)$ then IVP $y(x_0)=y_0$ has at least one solution on $I:a,b$ that contains $x_0$

Rephrase: A unique solution to $y^{\prime }$ and initial value $y(x_0)=y_0$ exists on some subinterval of $x$ when $f,f_y$ are continuous in an open rectangle $a<t<b$, $c<y<d$ containing initial point $(x_0,y_0)$

A unique solution to $y^{\prime }$ and initial value/conditions $(x_0,y_0)$ exists on some interval of $x$ denoted $I$, if functions $p,g$ are continuous on $I$ for DE $y^{\prime }+p(x)=g(x)$ and $I$ contains initial point $(x_0,y_0)$.

Typically if there are multiple solutions available (say $\pm$) we want to take the solution always containing the initial value $(x_0,y_0)$

Uniqueness §

Additionally if $f,f_y$ are continuous on $R$ then the IVP has a unique solution on some open subinterval of $(a,b)$ which contains $x_0$.

https://math.libretexts.org/Courses/Cosumnes_River_College/Math_420%3A_Differential_Equations_(Breitenbach)/01%3A_Introduction/1.02%3A_Existence_and_Uniqueness_of_Solutions