1. ## DIfferential equations

Solve each of the following differential equations to find the corresponding growth function. (show all necessary calculus)

i. dQ/dt=kQ where k>0 t>= 0 Q(0)= Qo

2. A koala population N is subject to a birth rate a and a deah rate of b+y p per head of population per annum. Ecologists have modelled the rate of change of this population over time using the following:
dN/dt = (a-b)N-yN^2

It is knowln that a = 0.2, b=0.04, y= 2x10^-4 and the initial population was 200.
Determine if and when the population will have doubled in size.

I have this question down to t= int(1/(0.16N-2*1-^-4N^2) dn. I can't seem to integrate it properly. If someone could integrate and point me in the right direction for the rest of the question that'd be great.

Another derivitives question for a homework sheet.

The ends of a 4m long water trough are equilateral triangles whose sides are a metre in length.
If water is being pumped into the trough at a rate of 1.5m^3 / min find the rate at which the water level is rising when the depth is 30 cm.

2. Originally Posted by phillipvdw2001
2. A koala population N is subject to a birth rate a and a deah rate of b+y p per head of population per annum. Ecologists have modelled the rate of change of this population over time using the following:
dN/dt = (a-b)N-yN^2

It is knowln that a = 0.2, b=0.04, y= 2x10^-4 and the initial population was 200.
Determine if and when the population will have doubled in size.

I have this question down to t= int(1/(0.16N-2*1-^-4N^2) dn. I can't seem to integrate it properly. If someone could integrate and point me in the right direction for the rest of the question that'd be great.
$\displaystyle t=\int_{200}^{400} \frac{1}{(a-b)N-2. 10^{-4}N^2} ~dN = \int_{200}^{400} \frac{1}{N[(a-b)-2. 10^{-4}N]} ~dN$

Now use partial fraction on the integrand to turn it into something like:

$\displaystyle \frac{A}{N}+\frac{B}{(a-b)-2. 10^{-4}N}$

RonL

3. Hello, phillipvdw2001!

Solve the differential equation to find the corresponding growth function.

. . $\displaystyle \frac{dQ}{dt}\:=\:kQ$ . where $\displaystyle k > 0,\;t \geq 0,\;Q(0) = Q_o$
Separate variables: .$\displaystyle \frac{dQ}{Q} \:=\:k\,dt$

Integrate: .$\displaystyle \ln Q \:=\:kt + c$

Solve for $\displaystyle Q\!:\;\;Q \:=\:e^{kt+c} \:=\;e^{kt}\cdot e^c\quad\Rightarrow\quad Q(t)\:=\:Ce^{kt}$

Since $\displaystyle Q(0) \,=\,Q_o$, we have: .$\displaystyle Q_o \:=\:Ce^0\quad\Rightarrow\quad C \,=\,Q_o$

Therefore: .$\displaystyle Q(t) \:=\:Q_oe^{kt}$