A multiple choice question concerning first order linear differential equations Consider the initial value problem $$y'(t)=f(t)y(t)$$ with $y(0)=1$ where $f:\mathbb{R}\rightarrow \mathbb{R}$ is a  continuous function. 
Then this initial value problem has 
(1) Infinitely many solutions for some $f$
(2)a unique solution in $\mathbb{R}$
(3) no solution in $\mathbb{R}$ for some $f$
(4) a solution in an interval containing $0$, but not on $\mathbb{R}$ for some $f$
My efforts 
$$\frac{dy(t)}{y(t)} = f(t)dt$$ 
$$\log y(t) = \int_0^tf(t)dt +C$$ 
Now at $t=0$ $y=1$ so we get 
$$0=0+C$$
So $$y(t)=\exp(\int_0^tf(t)dt)$$ 
So either $2$ is true or $4$ is true? I am not able to go further from here. What logic should I use now
Edit:
Since $\int_{0}^{t} f$ is continuous and $e^x$ is continuous and composition of continuous function is continuous so $2$ is true. Is this a correct way of thinking?
 A: I think the correct choice is (2) and only (2).
I reason as follows:
For $f(t) \in C(\Bbb R, \Bbb R)$, consider the function
$F_f: \Bbb R \times \Bbb R \to \Bbb R, \; F_f(y, t) = yf(t); \tag 1$
then $F_f(y, t)$ is (a.) jointly continuous in $y$ and $t$, and (b.) (locally in $t$) Lipschitz continuous in $y$.
To see (a.), note that $F_f(y, t) = yf(t)$ is the product of the two continuous functions $y: \Bbb R \to \Bbb R$ and $f: \Bbb R \to \Bbb R$, and that multiplication is itself continuous from $\Bbb R \times \Bbb R \to \Bbb R$.
As for (b.), we have, for $y, z, t \in \Bbb R$,
$\vert F(z, t) - F(y, t) \vert = \vert zf(t) - yf(t) \vert = \vert(z - y)f(t) \vert = \vert z - y \vert \vert f(t) \vert; \tag 2$
now since $f: \Bbb R \to \Bbb R$ is continuous, in any closed interval $I_{t_0}$ about any $t_0 \in \Bbb R$, $f(t)$ is bounded there; thus, there is a neighborhood $J_{t_0} \subset I_{t_0}$ of $t_0$ and a constant $L_J > 0$ such that
$\vert f(t) \vert < L_J, \; t \in J_{t_0}; \tag 3$
thus for $z, y \in \Bbb R$ and $t \in J$ we have
$\vert F(z, t) - F(y, t) \vert < L_J \vert z - y \vert, \tag 4$
which shows that $F(y, t)$ is Lipschitz continuous on some interval $J_{t_0}$ containing $t_0$; thus, by the Picard-Lindeloef theorem the exists a unique solution to our differential equation
$\dot y = F(y, t) = f(t)y, \; y(t_0) = y_0, \tag 5$
on some interval $(t_0 - \epsilon, t_0 + \epsilon)$ containing $t_0$; furthermore, since this is true for any $t_0$, such integral curves may be maximally extended to all of $\Bbb R$ and thus we see there exists a unique, global solution passing through any $(y_0, t_0) \in \Bbb R \times \Bbb R$; since this is true for any $f(t) \in C(\Bbb R, \Bbb R)$, we may rule out possibilities (1) and (3) in our OP Stammering Mathematician's list.  Similarly, we may eliminate item (4) from consideration since, as we have seen, there is a solution for every $f$ in an interval about any $t_0$, $t_0 = 0$ included.
So, (2) is true and only (2) is true.
The particularly simple form of $F(y, t) = f(t)y$ in fact makes it possible to verify the above results in an immediate, practical way without resorting to the theoretial arguments presented above; indeed, if
$y(t_0) = y_0 \ne 0 \tag 6$
for some $t_0 \in \Bbb R$, then in a neighborhood of $t_0$ we have
$\dfrac{\dot y(t)}{y(t)} = f(t), \tag 7$
whence
$\dfrac{d\ln y(t)}{dt} = f(t), \tag 8$
so that
$\ln \left (\dfrac{y(t)}{y(t_0)} \right ) = \ln y(t) - \ln y(t_0) = \displaystyle \int_{t_0}^t \dfrac{d\ln y(s)}{ds} \; ds = \int_{t_0}^t f(s) \; ds, \tag 9$
and so
$y(t) = y(t_0) \exp \left (\displaystyle \int_{t_0}^t f(s) \; ds \right ). \tag{10}$
The solution (10) is easily seen to be valid, unique, and to extend to all $\Bbb R$, and binds as long as $y(t_0) \ne 0$ for some $t_0 \in \Bbb R$; the only other alternative is $y(t) = 0$ everywhere, which is also seen to be a unique global solution.
Finally, if we choose with our OP
$y(0) = 1, \tag{11}$
we find the unique global solution to be
$y(t) = \exp \left ( \displaystyle \int_0^t f(s) \; ds \right ), \tag{12}$
in accord with our previous calculations.
A: You are correct. Since the integral, $\int _0^t f(t)dt $ exists for all t, you have a solution for all t.   
