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

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

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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 numbersare constants whose precise value will depend on the situation being modelled

      • In an exam question the values of the constants will generally be given to you

  • 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

  • It is usually more convenient, however, to 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

    • Here 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, and bold italic x equals 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?

  • The exact solution of the coupled system bold x with bold dot on top equals bold italic M bold italic x depends on the eigenvalues and eigenvectors of the matrix of coefficients bold italic M equals open parentheses table row a b row c d end table close parentheses

    • The eigenvalues and/or eigenvectors may be given to you in an exam question

    • If they are not then you will need to calculate them using the methods learned in the matrices section of the course

  • On the 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

    • Similar solution methods exist for non-real, non-distinct and/or non-zero eigenvalues, but you don’t need to know them as part of the IB AI HL course

  • Let the eigenvalues and corresponding eigenvectors of matrix bold italic M be lambda subscript 1 and lambda subscript 2, and bold italic p subscript 1 and bold italic p subscript 2, respectively

    • Remember from the definition of eigenvalues and eigenvectors that this means that bold italic M bold italic p subscript 1 equals lambda subscript 1 bold italic p subscript 1 and bold italic M bold italic p subscript 2 equals lambda subscript 2 bold italic p subscript 2

  • 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

    • This solution formula is in the exam 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 (see the worked example below)

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.

5-7-1-ib-ai-hl-solving-coupled-diff-eqns-we-solution

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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.