Differential equation of the form: y'(t) = ay(t) I stumbled upon this differential equation:

And here is the answer as stated: 

I don't really get how the answer is acquired and it is the first time for me to see a similar question.
Anyhow, thanks in advance!
 A: An intuitive solution to this equation is to think about what it's asking. First, look at the simpler equation $y^\prime = y$. If $y$ is a function, and it's derivative is the same as the original function, $y$ has to be $e^t$. Actually, $y$ can be any constant times $e^t$, since constants get factored out of derivatives.
Now, your equation says $y^\prime = ay$, where $a$ is some number. You started with a function $y$, and when you took a derivative, you got the original function, times a number. That sounds like a chain rule, where you take the derivative and multiply by the derivative of the inside. In this case, the derivative of the function is itself, and the derivative of the inside is $a$, so the inside should be $ax$. Thus, $y = e^{ax}$ times any constant will solve the equation. If we let $C$ be the constant, we can look at the solution for $t=0$ to see that $y(0) = C\cdot e^{a(0)}= C$, so the solution is $y(t) = y(0)\cdot e^{at}$.
A: $y=0$ is a solution.
if $y\neq 0$ in a certain interval, then
$$\frac {y'(t)}{y(t)}=a $$
and
$$\ln (\frac {y (t)}{\lambda})=at$$
You can finish using
$$e^x=1+x+\frac {x^2}{2!}+... $$
A: There are a few ways of looking at this.
One is to note that $y$ satisfies the equation iff
$t \mapsto e^{-at} y(t)$ is a constant function.
Another is to note that, if we assume that $y$ has a valid Taylor series
expansion (it does) then $y'(0) = ay(0), y''(0) = a^2 y(0),...$, etc. Then
$y(t) = y(0)\sum_{k=0}^\infty {(at)^k \over k!} = e^{at} y(0)$.
