Let's consider the initial value problem
$$ y^{\prime}(t) = f(t)y(t), y(0)=1 $$
where, $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, Then the IVP has
(a) infinitely many solutions for some $f$
(b) a unique solution in $\mathbb{R}$
(c) no solution in $\mathbb{R}$ for some $f$
(d) a solution in an interval containing $0$ but not on $\mathbb{R}$ for some $f$
We have to find the correct options. Can we do it like:
$$ y^{\prime}(t) = f(t)y(t)$$
$$y=c e^{\int f(t)dt}$$
$$y=c e^{g(t)}; \int f(t)dt=g(t) $$ on putting initial condition, we can find the value for $c$. In that case,can we reject options a and c. Any help, Please