theorem of existence and uniqueness for first order linear differential equation The theorem of existence and uniqueness is: 
Let $ y'+p(x)y=g(x) $, $ y(x_{0})=y_{0} $ be a first order linear differential equation such that $ p(x) $ and $ g(x) $ are both continuous for $ a<x<b $. Then there is a unique solution that satisfies it.
When a differential equation has no solution that satisfies $ y(x_{0})=y_{0} $, what does this mean?? Can the theorem be verified??
 A: 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 $y' + P(x)y = Q(x)$?
