Ordinary Differential Equations

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Ordinary Differential Equations

Subject:	Applied Mathematics
Question	10
Level	Honours only

[edit] Introduction

On the Applied Mathematics course an Ordinary Differential Equation (ODE) is defined as a relation that contains functions of only one independent variable, and one or more of its derivatives with respect to that variable.

An example of a typical question that a student might be given as an exercise is:(where a student would be asked to solve the equation for an expression in $y$ and $x$ only.)

$\Large x\Large y\frac{dy}{dx}=\(5+y^2)\$

[edit] Topic Structure

There are three types of Ordinary Differential Equations on the Leaving Certificate course that a student must be familiar with. These are:

First Order Differential Equations

A First Order Differential Equation has, as it's highest coefficient, a first derivative.

It's general form is:

$\frac{dy}{dx}=\Large y$

Second Order Separable Differential Equations

A Second Order Separable Differential Equation is an ODE that can be solved by replacing the second derivatives with first derivatives or functions.

It's general form is:

$\frac{d^2y}{dx^2}=f\frac{dy}{dx}$

Second Order Differential Equations that require the Chain Rule

A Second Order Separable Differential Equation is an ODE that can be solved by replacing the second derivatives with appropriate functions, or other derivatives, in such a way that the equation can be solved in two steps. Firstly by finding a value for the equation in terms of the newly entered function or derivative, and then by reviewing the original question with the new answers and therefore solving it completely.

It's general form is:

$\frac{d^2y}{dx^2}=f$

[edit] Applications to Power, Time, Forces or Distance

All questions that a student will encounter as a Part B of an exam paper will be applied to either Power, Time, Forces or Distance or a combination of these options. Students need to be particularly aware of the formulas that apply to each of these topic areas. These questions are always focused on acceleration and some manipulation on it - for example, a very common theme that has been examined in past exams is the rate of change of the acceleration of an object in free fall, due to the increasing air resistance experienced by it.

When solving these question students need to be able to replace acceleration with an appropriate derivative. The derivatives used are one of the following:

The rate of change of velocity $v$ , with respect to time $t$

$\Large a=\frac{dv}{dt}$

The rate of change of displacement $s$ , with respect to time $t$ squared

$\Large v=\frac{ds}{dt}$

$\int\Large v\ dt=\int\frac{ds}{dt}\ dt$

$\frac{dv}{dt}=\frac{d^2s}{dt^2}$

$\Large a=\frac{d^2s}{dt^2}$

Chain Rule applied to the rate of change of velocity $v$ , with respect to time $t$

$\Large a=\frac{dv}{dt}=\frac{dv}{ds}.\frac{ds}{dt}=\frac{dv}{ds}\Large v$

$\Large a=\Large v\frac{dv}{ds}$

[edit] Exam Paper

On the Honours exam paper, question 10 is the 'Differential Equations' question. The question is generally separated into two parts: Part A and Part B.

Part A consists of an Initial Value Problem (IVP), where the students are asked to find the original equation before differentiation or alternatively to find another value for $x$ or $y$ , given a value of $x$ or $y$ . These questions are usually not very difficult and only simple manipulation is needed to solve them.

Part B consists of an applied question to either Power, Time, Forces or Distance or a combination of those options. Students will have to find an expression for the acceleration and then answer a specific question regarding distance, time or velocity. These questions can be very difficult and it is primarily because of the general difficulty of this part of the question that the average marks are generally quite low in question 10.

The following worked example is taken from the 1999 exam paper:

10.(a) Solve the differential equation:
                                               
       Given that  when .

Solution:

The variables are first organised: all of one variable on the left hand side and all of the other variable on the right hand side.

$\left ( \frac{7}{v^2+1} \right )\frac {dv}{dx}=\frac {1}{x}$

$7\frac {dv}{v^2+1}=\frac {dx}{x}$

Both sides are then differentiated.

$7\int\frac {1}{v^2+1}\ dv\ =\int\frac {1}{x}\ dx$

$7\tan^{-1}\ v\ =\log_{e} x\ +\ C$

We now use or initial values to find $C$ , the constant of integration. We find that $C$ is zero. Therefore all that is needed now is to find the original equation.

$7\tan^{-1}\ v\ =\log_{e} x$

$tan^{-1}\ v\ =\frac {1}{7}\ log_{e} x$

$v\ =\tan \ \left (\frac {1}{7}\log_{e} x\right )$

10.(b) The rocket engine of a 12 tonne missile produces a thrust of 180.1kN.
       The missile is launched in a vertical direction.
       The air resistance is  N, where v is the speed of the missile.
       (i)   Find the speed of the missile after 30 seconds.
       (ii)  Find the percentage error in this speed if the air resistance is ignored.

Solution: Part (i)

For the first part of the question, the only formula that a student must be familiar with is Newton's Second Law:

$F=ma$

Concerning the rocket, three distinct forces are acting on it:

The thrust, which is 180,100 N and which will act vertically upwards and is positive.

The force $v^2$ N, which is a retarding force and will act downwards and is negative.

The weight of the rocket $mg$ which will be taken as a retarding force and will therefore act downwards and is negative.

Therefore, where $F$ is the force acting on the rocket:

$F=\ 180,100\ -\ v^2\ -\ mg$

$F=\ 180,100\ -\ v^2\ -\ 117,600\ =\ 62,500\ -\ v^2$

$a=\ \frac {62,500-v^2}{12,000}$

Now that we have an expression for the acceleration, we can begin the question proper.

Since the first part of the question requires time and velocity, we will let:

$\Large a=\frac{dv}{dt}$

$\frac{dv}{dt}\ =\ \frac {62,500-v^2}{12,000}$

$\frac {12,000}{62,500-v^2}\ dv=\ dt$

$12,000\int \frac {1}{62,500-v^2}\ dv=\ \int dt$

$12,000 \left (\frac {1}{500} \log_{e} \left |\frac {250+v}{250-v}\right |\right )\ =\ t\ +\ C$

$24\left (\log_{e} \left |\frac {250+v}{250-v}\right |\right )\ =\ t\ +\ C$

We now use or initial values to find $C$ , the constant of integration. Our initial values are that the velocity $v$ is zero when the time $t$ is zero - since the rocket takes off from a stand-still.

We find that $C$ is zero. Therefore:

$24\left (\log_{e} \left |\frac {250+v}{250-v}\right |\right )\ =\ t$

We need to find the value of the velocity $v$ , when the time $t$ is 30 seconds.

$\log_{e} \left |\frac {250+v}{250-v}\right |\ =\frac {30}{24}$

We only consider the positive value of e.

$e^{\frac{5}{4}}\ = \left (\frac {250+v}{250-v}\right )$

Simple manipulation yields:

$v\ =\ 138.649\ m/s$

Solution: Part (ii)

If we disregard air resistance, we can see that the acceleration is a constant and therefore no integrals are to be found.

$a\ =\frac {62,500}{12,000}\ =\frac {125}{24}$

$v\ =\ u\ +\ at$

$v\ =\ 0\ +\ \frac {125}{24}.(30)\ =\ 126.25$

Therefore the percentage error is simply the the difference between the two, divided by the correct velocity, which yields an answer of 12.69%.

[edit] Criticisms

On two separate exam papers, the 'Differential Equations' question has included some part of a question that is not on the prescribed curriculum as layed out by the Department of Education.

On the 2001 paper, part A of question 10 featured a non-separable ODE, and is therefore by definition, not on the prescribed curriculum.

The 1999 paper (which is used as a worked example above) featured in its part B of question 10, a logarithmic integral which is not studied as part of the prescribed Mathematics curriculum and is therefore also not on the prescribed curriculum for Applied Mathematics.

Chief Examiners Reports for those years in question are at the moment unavailable on the Examinations Commission website, so the commissions stance on the inclusion of these questions is largely unknown.

Who Added These Notes?

Zorba

Ordinary Differential Equations

From ZuluNotes - Free Leaving Cert Notes

Contents

[edit] Introduction

[edit] Topic Structure

[edit] Applications to Power, Time, Forces or Distance

[edit] Exam Paper

[edit] Criticisms

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Ordinary Differential Equations

From ZuluNotes - Free Leaving Cert Notes

Contents

[edit] Introduction

[edit] Topic Structure

[edit] Applications to Power, Time, Forces or Distance

[edit] Exam Paper

[edit] Criticisms

Who Added These Notes?

Views

Personal tools

Navigation

Search

Subjects

Toolbox

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