Investment approach infinity We know that the value of an investment, $y(t)$, which is compounded continusouly at the interest rate k is governed by the equation $\frac{dy}{dt} = ky$. If the interest rate is related to the size of the investment by $k = \frac{\sqrt(y)}{60}$ then we have the equation $\frac{dy}{dt} = \frac{y^{3/2}}{60}.$ a) If one initially invests $400 what is the value of your investment at later times? (it looks like this is an equation not a specific number). b) For a given initial investment, $y_0$, determine the time at which the worth of the investment approaches !.

So you have the differential equation

$$y'(t) = a y^{3/2}$$

where $a=1/60$, with $y(0)$ being given. To solve:

$$y^{-3/2} dy = a dt \implies -2/\sqrt{y} = a t + C$$

where $C=-2/\sqrt{y(0)}$.

Then we have

$$\frac{2}{\sqrt{y(t)}} = \frac{2}{\sqrt{y(0)}}-a t$$

This goes to $\infty$ when

$$t=\frac{2}{a \sqrt{y(0)}} = \frac{120}{\sqrt{y(0)}} = 6$$

**EDIT**

This answers (b), which was the original question. (a) asks for the form of $y(t)$ when $y(0)=400$. Doing the algebra on my previous equation, I get

$$y(t) = \left ( \frac{1}{20}-\frac{1}{120} t\right)^{-2} = \frac{400}{\left (1-\displaystyle\frac{t}{6}\right)^2}$$