Differential equations. Shortcut way to solve this problem? Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.
Here is the problem I am trying to solve:

The weight in pounds of a certain bear cub $t$ after birth is given by
  $w(t)$. If $w(2)=36, w(7)=84,$ and $\frac{dw}{dt}$ was proportional to
  the cub's weight for the first $15$ months of his life, how much did
  the cub weigh when he was $11$ months old?


A friend of mine emailed me his solution:

No calculus involved. $\frac{dy}{dy} = k y$ That implies $y=\exp(kt+C)$ One $(t,y)$
  pair gives $k$, another gives $C$. You don't have to know the
  differentiation, just the result, hence, no Calculus. 

Can someone decipher what he's saying?  What is $\exp()$?  Exponent?  $y$ equals an exponent?  Huh?  I had no idea what he meant, and asked for clarification.    His alternate solution was 

If Diff. Eq. is of the form $\frac{dy}{dt} = ky$, then write solution as $y=$
  $\exp(...)$ That's it!

I don't understand how this problem can be solved in just 1 line.  It's clear that he's addressing this problem with a totally different approach than the traditional "Calculus/DiffEq" approach.  
Below is is how I did it.   Although I arrive at the correct answer after 10 mins. and an entire page of paper, I'd like to understand the 1-liner shortcut method above, as it seems a big time saver.  Can anyone explain his approach written out legibly in a photo scan?

 A: Better to write this way: if $\dot w = k w,$ then $$  w = C e^{k t}.  $$ You do not really need to solve for $k,$ because
$$  w = C \left( e^k \right)^t. $$ So, if we define a positive real number $M = e^k,$ we get
$$  w = M^t.   $$ 
This one does not come out pretty. I guess you are taking an approximation.
Given $w(2)= 36, w(7) = 84,$ we have
$$ C M^2 = 36, \; \;  C M^7 = 84,  $$ and
$$  M^5 = \frac{84}{36} =  \frac{7}{3}, $$
so
$$ M =   \sqrt[5]\frac{7}{3} \approx 1.184664 , $$ meanwhile
$$ C = 36 / M^2 \approx 25.65144.  $$
$C$ is the weight at birth.
Oh, well. Weight at 11 is 
$$ C M^{11} = C M^7 \cdot M^4 = 84 \cdot M^4 \approx 84 \cdot 1.969615 \approx 165.44769  $$
Maybe this will help: because $7-2 = 5,$ we get nice rational number weights at $t=2,7,12,17,22,$ and so on. 
$$  C M^{12} = 84 \cdot M^5 = 84 \cdot \frac{7}{3} = 196, $$
$$  C M^{17} = 196 \cdot M^5 = 196 \cdot \frac{7}{3} = \frac{1372}{3} = 457 + \frac{1}{3}. $$
The one with the 196 gives a way to confirm the answer, 
$$  C M^{11} = \frac{C M^{12}}{M} = \frac{196}{M} \approx  \frac{196}{1.184664} \approx 165.44769   $$
A: Imagine two cubs starting with weights $w_0$ and $w_1$. When for both of them ${dw\over dt}=c\, w(t)$  with the same proportionality constant $c$, then the weights of both of them  multiply by the same factor within a week. From the data we conclude that this factor is $\left({84\over36}\right)^{1/5}$. Therefore in the remaining four weeks the cub increases his weight to $\left({84\over36}\right)^{4/5}\cdot 84=165.448$ pounds.
