Consider the IVP;
$\text{y$^{'}$(t)=f(t)y(t),y(0)=1}$;
where $f$ is continuous.
Show that the IVP has a unique solution in $\Bbb R$.
Since $\text{y$^{'}$(t)=f(t)y(t)}\implies \dfrac{y^{'}(t)}{y(t)}=f(t)\implies \ln y=f(t)\implies y(t)=e^{f(t)}+C\implies 1=e^{f(0)}+C\implies C=1-e^{f(0)}$.
Then $y(t)=e^{f(t)}+(1-e^{f(0)})$.
But I am not sure where to use that $f$ is continuous and how to show that solution is unique?
I am helpless.Please help me.