Consider a first order linear differential equation $y'+Py=Q,\,y(a)=y_0$ .Here continuity of $P$ and $Q$ ensure that the the ODE has an unique solution.But in case of non-linear initial value problem i.e $y'=f(x,y),\,y(x_0)=y_0$,continuity of $f$ does not ensure the unique solution.So we need to address the following question:
(1)Under what condition on $f$ the problem $y'=f(x,y),\,y(x_0)=y_0$ has a solution$?$
(2)If solution exists,whether it is unique or not$?$
$\to$First question is answered by the Peano existence theorem which states that "let $f$ be a continuous function in an interval $I$ containing the points $(x_0,y_0)$,then the the problem $y'=f(x,y),\,y(x_0)=y_0$ has a solution".
$\to$Second question is answered by Picard's uniqueness theorem which states that "let $f$ and $\frac{\delta f}{\delta y}$ are continuous in aregion R containing the initial points $(x_0,y_0)$ then the the problem $y'=f(x,y),\,y(x_0)=y_0$ has an unique solution"
Picard method for interval of definition:let $f$ and $\frac{\delta f}{\delta y}$ are continuous in a closed rectangle $$R=\{(x,y):|x-x_0|\leq a,|y-y_0|\leq b\}$$.Then the IVP $y'=f(x,y),\,y(x_0)=y_0$ has an unique solution in the interval $|x-x_0|\leq h=min{(a,\frac{b}{l})}$ where $l=MAX_{(x,y)\in R}|f(x,y)|$
Hope this will help you!!!
Best Answer
For Nagumo's theorem see
https://projecteuclid.org/download/pdf_1/euclid.pja/1265033221
and
https://www.jstor.org/stable/pdf/2036312.pdf
Foe Osgood's theorem see
http://www.math.toronto.edu/mpugh/Teaching/MAT267_19/Osgood_Uniqueness_Theorem.pdf