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I am claiming that: Let $p(t)$ be continuous and $q(t)$ be integrable function on $[a,b]$. Then uniqueness and existence are also valid for $$y^\prime+p(t)y=q(t), \;\;\;\; y(t_0)=a_0$$ where $t_0 \in [a,b]$. I checked some proofs, all of them prove for that $q$ is continuous. Am I missing something?

We need continuity of $p$ since we need differantibility of $$\int_{t_0}^tp(s)ds.$$ (if we want to use Fundamental Theorem of Calculus). We are just using $q$ for saying that $pq$ is integrable. It is enough that ($\mu(t)=e^{\int_{t_0}^tp(s)ds}$) $\mu q$ is integrable, since $\mu$ is (continuous so) integrable on $[a,b]$ and product of integrable functions is again integrable.

For uniqueness part, we actually don't use $q$.

My second question is that: if there is more weaker conditions, what will be them?

Thanks in advance.

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    Actually, if you require the solution to be $C^1$, you notice that $q$ has to be continuous because sum of continuous functions. – 2017-02-08
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    By FTC, if $h$ is integrable on $[a,b]$, then $H(t)=\int_a^t h(s)ds$ gives a continuous function H on $[a,b]$. We do not need continuity. – 2017-02-09
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    By the way, $y^\prime$ has not be continuous. – 2017-02-09

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Since you say in the comments that $y$ need not be $C^1$, we need first to agree what is the meaning of "solution". I have to presume that it is in the weak/mild sense, that is, a solution should be a continuous function $y$ on some open interval containing $t_0$ such that $$ y(t)=a_0+\int_{t_0}^t(-p(s)y(s)+q(s))\,ds $$ for all $t$ in the interval. This would be the most general notion without looking say for solutions that are distributions.

In this setting there exists a solution even if $p$ and $q$ are measurable. Look for example at Hale's ODE book for the Carathéodory conditions, which are among the most general leading to at least one weak/mild solution.

For the uniqueness you don't need continuity, but only something that leads to the local Lipschitz property and then just apply the usual Gronwall argument. For example, it suffices that $p$ is bounded on compact sets.

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    Thanks for answer. I will check the book. I actually said it for $y^\prime$. – 2017-02-10
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    Note that $y'$ is continuous if and only if $y$ is $C^1$. – 2017-02-10
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    Oops, I misunderstood the notation. I supposed C^1 is for continuous function, I checked it now, it is for differantible functions. – 2017-02-10