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 !. 
 A: If you are continuously compounding interest at a nominal rate of $k$ per annum, you are correct that the equation is $\frac {dy}{dt} = ky$.  If the interest rate is fixed, the solution is $y=y(0)\exp(kt)$.  This never goes to infinity in finite time.  If you are given $k=\frac 1{60}\sqrt y$, the equation becomes $\frac {dy}{dt}=\frac 1{60}y^{\frac 32}$.  Integrating this we get $$\int y^{-\frac 32} dy=\int \frac 1{60} dt \\
-2y^{-\frac 12}=\frac t{60}+C\\
\frac 1{\sqrt y}=-\frac C2-\frac t{120}\\\sqrt y=-\frac 1{\frac C2+\frac t{120}}$$ Since $y \gt 0$, we must have $C \lt 0,$ so we will put $C'=-C$ to get $$\sqrt y=\frac 2{C'-\frac t{60}}\\\sqrt y=\frac {120}{60C'-t}$$  At $t=0$ we have $C'=\frac 2{\sqrt {y(0)}}$ So we have $$\sqrt y=\frac {120}{\frac {120}{\sqrt {y(0)}}-t}$$ which becomes infinite at $$t=120y(0)^{-\frac 12}$$
A: 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
$$y(t) = \left ( \frac{1}{20}-\frac{1}{120} t\right)^{-2} = \frac{400}{\left (1-\displaystyle\frac{t}{6}\right)^2}$$
