IBMathsDPApplications & Interpretation (AI)HLRevision Notes5. CalculusCoupled & Second Order Differential EquationsCoupled Differential Equations

Coupled Differential Equations (DP IB Applications & Interpretation (AI)): Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

DP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): HL

Solving coupled differential equations

How do I write a system of coupled differential equations in matrix form?

  • The coupled differential equations considered in this part of the course will be of the form

    fraction numerator d x over denominator d t end fraction equals a x plus b y

    fraction numerator d y over denominator d t end fraction equals c x plus d y

    •  a comma space b comma space c comma space d space element of space straight real numbers are constants whose precise value will depend on the situation being modelled

  • This system of equations can also be represented in matrix form:

open parentheses table row cell fraction numerator d x over denominator d t end fraction end cell row cell fraction numerator d y over denominator d t end fraction end cell end table close parentheses equals open parentheses table row a b row c d end table close parentheses open parentheses table row x row y end table close parentheses

  • You can use the dot notation for the derivatives:

open parentheses table row cell x with dot on top end cell row cell y with dot on top end cell end table close parentheses equals open parentheses table row a b row c d end table close parentheses open parentheses table row x row y end table close parentheses

  • This can be written even more succinctly as bold italic x with bold dot on top bold equals bold italic M bold italic x

    • bold italic x with bold dot on top equals open parentheses table row cell x with dot on top end cell row cell y with dot on top end cell end table close parentheses

    • bold italic M equals open parentheses table row a b row c d end table close parentheses

    • bold italic x equals open parentheses table row x row y end table close parentheses

  • For example,

    • fraction numerator d x over denominator d t end fraction equals 2 x minus 3 y and fraction numerator d y over denominator d t end fraction equals x plus 4 y can be modelled as

    • open parentheses table row cell x with dot on top end cell row cell y with dot on top end cell end table close parentheses equals open parentheses table row 2 cell negative 3 end cell row 1 4 end table close parentheses open parentheses table row x row y end table close parentheses

How do I find the exact solution for a system of coupled differential equations?

Examiner Tips and Tricks

In your exam, you will only be asked to find exact solutions for cases where the two eigenvalues of the matrix are real, distinct, and non-zero.

  • Suppose for matrix bold italic M

    • lambda subscript 1 and lambda subscript 2 are the eigenvalues

    • bold italic p subscript 1 and bold italic p subscript 2 are corresponding eigenvectors respectively

  • The exact solution to the system of coupled differential equations is then

    bold italic x equals A straight e to the power of lambda subscript 1 t end exponent bold italic p subscript 1 plus B straight e to the power of lambda subscript 2 t end exponent bold italic p subscript bold 2

Examiner Tips and Tricks

This is given in your formula booklet. A comma space B space element of space straight real numbers are constants. They are essentially constants of integration of the sort you have when solving other forms of differential equation.

  • If initial or boundary conditions have been provided you can use these to find the precise values of the constants A and B

    • Finding the values of A and B will generally involve solving a set of simultaneous linear equations

Worked Example

The rates of change of two variables, x and y, are described by the following system of coupled differential equations:

fraction numerator d x over denominator d t end fraction equals 4 x minus y fraction numerator d y over denominator d t end fraction equals 2 x plus y  

Initially x equals 2 and y equals 1.

 

Given that the matrix open parentheses table row 4 cell negative 1 end cell row 2 1 end table close parentheses has eigenvalues of 3 and 2 with corresponding eigenvectors open parentheses table row 1 row 1 end table close parentheses and open parentheses table row 1 row 2 end table close parentheses, find the exact solution to the system of coupled differential equations.

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    Author
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    Roger B

    Author: Roger B

    Expertise: Maths Content Creator

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.