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IBMathsDPApplications & Interpretation (AI)HLRevision Notes5. CalculusCoupled & Second Order Differential EquationsSketching Solution Trajectories

Sketching Solution Trajectories (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): HL

Sketching solution trajectories

How do I sketch a solution trajectory for a system of coupled differential equations?

  • A phase portrait shows typical trajectories representing all the possible solutions to a system of coupled differential equations

  • If you are given a set of initial conditions then there will be one solution trajectory

  • To sketch a solution trajectory

    • Find the eigenvalues and eigenvectors of the matrix of coefficients (if needed)

      • They might be given in the question

    • Mark the starting point on the diagram

      • This will be given in the question

      • Or you will be given the solution and need to substitute t equals 0

    • Determine the initial direction of the trajectory

      • Find the values of fraction numerator d x over denominator d t end fraction when t equals 0 to determine if it is moving right or left

      • Find the values of fraction numerator d y over denominator d t end fraction when t equals 0 to determine if it is moving up or down

    • Sketch the trajectory using the general shape based on the eigenvalues

      • You should sketch lines for the eigenvectors if the eigenvalues are real

Worked Example

Consider the system of coupled differential equations

fraction numerator d x over denominator d t end fraction equals x minus 5 y fraction numerator d y over denominator d t end fraction equals negative 3 x plus 3 y

The initial conditions of the system are such that the exact solution is given by 

bold italic x equals straight e to the power of 6 t end exponent open parentheses table row cell negative 1 end cell row 1 end table close parentheses minus 2 straight e to the power of negative 2 t end exponent open parentheses table row 5 row 3 end table close parentheses

Sketch the trajectory of the solution, showing the relationship between x and y as t increases from zero.

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    Author
    Reviewer
    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.