Setting up and Solving Differential Equations Growth and Decay Objectives: 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 SOLVING If dy= y dx dx The equation tells us that the graph of y has a gradient that always equals 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 SOLUTIONS We call this family of solutions the general solution of a differential equation. We call this family of solutions the general solution of a differential equation.

Solving We use a method called “ Separating the Variables” and the title describes exactly what we do. We rearrange so that x terms are on the right and y on the left. Now insert integration signs... and integrate 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.

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 =... ) A log is just an index, so ( We now have the exponential that we spotted from the gradient diagram. ) However, it can be simplified.

So, We can write as. where k is positive This is usually written as where A is positive or negative. So, 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.

Changing the value of A gives the different curves we saw on the gradient diagram. e.g. A = 2 gives

Have a go: Solve the equation below giving the answer in the form Solution:Separating the variables: Insert integration signs: Integrate: Particular solution: X=½Π y=-½ => C=1

Setting up and Solving Differential Equations Growth and Decay Objective: To formulate and solve growth and decay problems involving proportional change.

Exercise 1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 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. ( When solving, use k correct to 3 s.f. )

Solution: Let t = 0 in The population of a town was 60,000 in 1990 and had increased to 63,000 by 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.

( nearest hundred ) 1. The population of a town was 60,000 in 1990 and had increased to 63,000 by 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: 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. (3 s.f.)

Substitute Ans: 5 hrs 44 mins 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.

 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  The solution to the above equation is ( 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, 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. Explain 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 by at time t minutes after it was poured. The temperature of the room in which the cup is placed is

We can solve this equation as follows: If we are given further information, we can complete the solution as in the other examples.