Differentiation Equations

1. Find in terms of given that .

 Hint

Solve by “separating variables” method: arrange the equation so that all terms involving are on one side and all terms involving are on the other side, then integrate both sides.

 Solutions

Separate variables:

Integrate both sides:

Exponentiate and solve for :

where is an arbitrary constant.

2. Solve for and , given that when .

 Solutions

Separate variables:

Integrate both sides:

Exponentiate (with ):

Use the initial condition :

Therefore the solution is

3. Obtain a particular solution to given that when .
   (There is no need to express in terms of ).

 Solutions

Separate variables:

Integrate both sides:

Use the condition :

Thus a particular (implicit) solution is

or equivalently .

4. Find an expression for in terms of given that .

 Hint

Try to find a way to separate the variables and . One way is to rearrange the equation to isolate , then rewrite it in differential form to separate the variables:

 Solutions

5. At time seconds the rate of increase in the concentration of flesh eating bugs in a controlled environment is proportional to the concentration of bugs present. Initially bugs and after 2 seconds there are five times as many.
   (a) Write down a differential equation connecting and and hence find an expression for in terms of .

 Hint

Since the rate of increase is proportional to the concentration:

where is a constant. Solve by separating variables method. Then use the initial conditions to find the constants.

 Solution

Since the rate of increase is proportional to the concentration:

where is a constant.

Separate variables and integrate:

Integral both sides:

so

Using the initial condition gives , hence

Now we need to find , Use to find :

Therefore

(b) How many bugs are present after 5 seconds?

 Solutions

When will the number of bugs exceed 5000 ?

 Solutions

We have . We want the time when .

Taking logarithms (or using base‑5 logs),

Equivalently (natural logs),

So the number of bugs exceeds 5000 after approximately seconds.

(d) Find the time at which the concentration of bugs has increased by of the initial concentration.

 Solutions

We have . We want the time such that the concentration has increased by of the initial value, i.e. .

6. Water is pouring out of a small hole in the bottom of a conical container of height 25 cm . Initially the container is full.
   The rate at which the height of the water remaining in the container is given by

(a) Solve the differential equation to find in terms of .

 Solutions

Separate variables:

Integrate:

Multiply both sides by :

Using the initial condition gives . Hence

Therefore

The container empties when

(b) How long does it take for the container to empty completely?

 Solutions

Setting gives

Numerically,

7. The rate at which a body loses temperature at any instant is proportional to the amount by which the temperature of the body, at that instant, exceeds the temperature of its surroundings.
   A cup of tea is in a room. The ambient temperature of the room is and the temperature of the tea is . minutes later the tea’s temperature has fallen by .
   (a) How long will it take for the tea to cool to ?

 Solutions

Separate variables and integrate:

Need to find :

We know that at , , so:

(General solution: )

So with , this becomes

Use the information at minutes: the temperature has fallen by , so . Thus

Find when :

(b) Suggest one limitation of this model.

 Solutions

It requires that the ambient temperature remains constant. In reality, the ambient temperature may vary over time.