# 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?

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The next time you solve a similar problem you can start with equation $w=Ce^{kt}$ and substitute the given conditions. –  Américo Tavares Feb 10 at 22:16
What is the rationale behind starting directly with that equation? I see no mention of continuous compounding or exponential growth/decay in the original problem. What rule of thumb/shortcut am I unaware of? –  JackOfAll Feb 10 at 22:28
Because the solution of $dw/dt=kw$ is $w=Ce^{kt}$ as you computed. If $k=1$, then $w=Ce^{t}$. And if $w(0)=1$, then $w=e^{t}$, the exponential function. –  Américo Tavares Feb 10 at 22:32
Wait, the solution of dw/dt=kw is $w=Ce^{kt}$? How? This is how I understand derivatives: If the derivative of w is kw, then isn't the antiderivative/original function $y=\frac{k}{2}w^2$ ?? –  JackOfAll Feb 10 at 22:56
$$w=Ce^{kt}\Rightarrow \frac{dw}{dt}=\frac{d}{dt}Ce^{kt}=C\frac{d}{dt} e^{kt}=Ce^{kt}k=kw.$$ –  Américo Tavares Feb 10 at 23:03

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, $$C M^{11} = \frac{C M^{12}}{M} = \frac{196}{M} \approx \frac{196}{1.184664} \approx 165.44769$$

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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.

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Clever! Clever! –  JackOfAll Jul 13 at 12:24