The fundamental theorem tells you that
$$
f(x,y)-f(x,z)=\int_{[y,z]}f_ydy=\int_0^1f_y(x,y+t(z-y))·(z-y)\,dt.
$$
The region $\{(x,y);|x−x_0|⩽a,\|y−y_0\|⩽b\}$ is a compact set, $f_y$ is continuous thus bounded, let $L$ be a bound. Then
$$
\|f(x,y)-f(x,z)\|\le\int_0^1L·\|z-y\|\,dt=L·\|z-y\|
$$
which provides the Lipschitz condition on $f$. Now apply Picard-Lindelöf.
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
The existence and uniqueness theorem for first-order linear differential equations can be stated as follows. Suppose that $P$ and $Q$ are continuous on the open interval $I$. If $a$ and $b$ are any real numbers, then there is a unique function $y = f(x)$ satisfying the initial-value problem $y' + P(x)y = Q(x)$ with $f(a) = b$ on the interval $I$. With regard to your question, the important point is that $a$ and $b$ are arbitrary real numbers and the unique solution $f$ to the differential equation satisfies $f(a) = b$ for every choice of $a$ and $b$. Since every first-order linear differential equation satisfying the constraints of the theorem has a solution satisfying $f(a) = b$, there is no case in which such an equation has no solution satisfying $f(a) = b$.
If we look at the simpler case of homogeneous first-order linear differential equations of the form $y' + P(x)y = 0$, where $P$ is continuous on the open interval $I$, we can directly verify that for every choice of $a$ and $b$, the function $f(x) = be^{-A(x)}$ where $A(x) = \int_{a}^x P(t) dt$ is a solution to $y' + P(x)y = 0$. Now letting $g$ be an arbitrary solution of $y' + P(x)y = 0$, we establish uniqueness by showing that $g(x)e^{A(x)} = b$. Differentiating, we see that $h(x) = g(x)e^{A(x)}$ is constant on the interval $I$. But $h(a) = b$, so we must have $h(x) = b$. This demonstrates that $g = f$. Notice that the choice of $a$ and $b$ does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form $y' + P(x)y = Q(x)$?