Setting up and Solving Setting up and Solving Differential Differential EquationsEquations

Growth and DecayObjectives:

To be able to find general and particular solutions to differential equations using the method of separating the variables.

To formulate and solve growth and decay problems involving proportional change.

BEFORE WE START SOLVINGBEFORE WE START SOLVING

If If dydy= y= y

dxdx

The equation tells us that the The equation tells us that the graph of graph of yy has a gradient that has a gradient that always equalsalways equals y y..

We can now draw a curve through any point following the gradients.

However, we haven’t got just one curve.

The solution is a family of curves.Can you guess what sort of equation these

curves represent ?ANS: They are exponential curves.

FAMILY OF SOLUTIONSFAMILY OF SOLUTIONS

We call this family of We call this family of solutions the general solutions the general solution of a differential solution of a differential equation.equation.

ydxdy Solvin

gWe use a method called “ Separating the Variables” and the title describes exactly what we do.

ydx

dydxdy

y

1

We rearrange so that x terms are on the right and y on the left.

Now insert integration signs . . .

dxdyy

1and integrate

Cxy ln

We can separate the 2 parts of the derivative because although it isn’t actually a fraction, it

behaves like one.

(the l.h.s. is integrated w.r.t. y and the r.h.s. w.r.t. x)

Multiply by dx and divide by y.

We don’t need a constant on both sides as they can be combined. I usually put it on the r.h.s.

ydx

dy

We’ve now solved the differential equation to find the general solution but we have an implicit equation and we often want it to be explicit ( in the form y = . . . )

Cxy ln

A log is just an index, so

Cxyln Cxey

( We now have the exponential that we spotted from the gradient diagram. )

However, it can be simplified.

So, xkeyydx

dy

We can write as .

Cxe Cx ee

where k is positiveThis is usually written as

where A is positive or negative. xAey

Cxey

So, ydx

dy xAey In this type of example, because the result is valid for positive and negative values, we usually use A directly when changing from log to exponential form.

Since is a constant it can be replaced by a single letter, k.

Ce

Changing the value of A gives the different curves we saw on the gradient diagram.

xAeyydx

dy

e.g. A = 2 gives

xydx

dycos2

Have a go: Solve the equation below giving the answer in the form

.)(xfy

Solution:

Separating the variables:

dxxdyy

cos1

2

Insert integration signs:

dxxdyy cos2

Cxy

sin1

1

Integrate:

Cxy

sin1

1sin

1

xy

Particular solution:

X=½Π y=-½

=> C=1

C1

1

2

1

Setting up and Solving Setting up and Solving Differential Differential EquationsEquations

Growth and DecayObjective:

To formulate and solve growth and decay problems involving proportional change.

Exercise1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred.2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute.

50

050

10

( When solving, use k correct to 3 s.f. )

nkdt

dn1.

( When solving, use k correct to 3 s.f. )

xkdt

dx 2.

Solution:

60000,0 nt

kteAn

60000A

63000,10 nt )10(6000063000 ke

Let t = 0 in 1990.

60000

63000ln10k

004880k

kndt

dn

1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred.

ten 00488060000

,60000A 004880k,kteAn

)20(0048806000020 ent

200,66 ( nearest hundred )

1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 2000. Assuming that the population is increasing at a rate proportional to its size at any time, estimate the population in 2010 giving your answer to the nearest hundred.

Solution:

50,0 xt 50A

10,4 xt

xkdt

dx ktAex

)4(5010 ke

50

10ln4k 4020k

2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute.

50

050

10

(3 s.f.)

tex 402050 :050xSubstitute

te 402050050

50

050ln4020 t

7285tAns: 5 hrs 44 mins

,ktAex 4020k,50A

2. A patient is receiving drug treatment. When first measured, there is mg of the drug per litre of blood. After 4 hours, there is only mg per litre. Assuming the amount in the blood at time t is decreasing in proportion to the amount present at time t, find how long it takes for there to be only mg. Give the answer to the nearest minute.

50

050

10

The words “ a rate proportional to . . . ” followed by the quantity the rate refers to, gives the differential equation for growth or decay.

SUMMARY

e.g. “ the number, x, increases at a rate proportional to x ” gives kx

dt

dxx

dt

dx

The solution to the above equation is

ktAex ( but if we forget it, we can easily separate

the variables in the differential equation and solve ).

either 1 pair of values of x and t and 1 pair

of values of and t,dt

dx

or 2 pairs of values of x and t.

The values of A and k are found by substituting

There is one very well known situation which can be described by a differential equation.

The following is an example.

Solution:The equation gives the rate of decrease of the temperature of the coffee. It is proportional to the amount that the temperature is above room temperature.

)20( xkdt

dxExplain what the following equation is describing:

This is an example of Newton’s law of cooling.

The temperature of a cup of coffee is given byat time t minutes after it was poured. The temperature of the room in which the cup is placed is

C20

Cx

)20( xkdt

dx

We can solve this equation as follows:

kdtdxx

20

1 Cktx 20ln

ktAex 20ktAex 20

If we are given further information, we can complete the solution as in the other examples.