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

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Euler's method: second order

What is a second order differential equation?

  • A second order differential equation is a differential equation containing one or more second derivatives

  • In this course you consider second order differential equations of the form

    fraction numerator straight d squared x over denominator straight d t squared end fraction equals f open parentheses x comma fraction numerator straight d x over denominator straight d t end fraction comma t close parentheses

  • Some examples include;

    • fraction numerator straight d squared x over denominator straight d t squared end fraction equals x fraction numerator straight d x over denominator straight d t end fraction plus 2 x squared plus 3 t plus 5

    • fraction numerator straight d squared x over denominator straight d t squared end fraction equals 2 fraction numerator straight d x over denominator straight d t end fraction plus 3

How do I apply Euler’s method to second order differential equations?

  • The pair of equations then becomes the system

fraction numerator d x over denominator d t end fraction equals y
fraction numerator d y over denominator d t end fraction equals f open parentheses x comma space y comma space t close parentheses

  • You can then use Euler's method in the same way

  • Write down the recursion equations using the formulae from the exam formula booklet:

    • x subscript n plus 1 end subscript equals x subscript n plus h cross times f subscript 1 open parentheses x subscript n comma blank y subscript n comma blank t subscript n close parentheses

    • y subscript n plus 1 end subscript equals y subscript n plus h cross times f subscript 1 open parentheses x subscript n comma blank y subscript n comma blank t subscript n close parentheses

    • t subscript n plus 1 end subscript equals t subscript n plus h 

      • h in those equations is the step size

      • The exam question will usually tell you the correct value of h to use

  • Use the recursion feature on your GDC to calculate the Euler’s method approximation over the correct number of steps

    • The values for x subscript 0, y subscript 0 and t subscript 0 will come from the boundary conditions given in the question

    • Frequently you will be given an initial condition

      • Look out for terms like ‘initially’ or ‘at the start’

      • In this case t subscript 0 equals 0

Worked Example

Consider the second order differential equation fraction numerator straight d squared x over denominator straight d t squared end fraction plus 2 fraction numerator straight d x over denominator straight d t end fraction plus x equals 50 cos invisible function application t.

a) Show that the equation above can be rewritten as a system of coupled first order differential equations.

5-7-2-ib-ai-hl-eulers-method-second-order-a-we-solution

b) Initially x equals 2 and fraction numerator d x over denominator d t end fraction equals negative 1. By applying Euler’s method with a step size of 0.1, find approximations for the values of x and fraction numerator d x over denominator d t end fraction when t equals 0.5 .

5-7-2-ib-ai-hl-eulers-method-second-order-b-we-solution

 

Exact solutions of second order differential equations

How can I find the exact solution for a second order differential equation?

  • You might be asked to find exact solutions to second order differential equations of the form

    fraction numerator straight d squared x over denominator straight d t squared end fraction plus a fraction numerator straight d x over denominator straight d t end fraction plus b x equals 0

    •  where a and b are constants

  • Use the substitution space y equals fraction numerator d x over denominator d t end fraction to turn it into a coupled differential equation

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

  • This can be written in the form bold italic x with dot on top equals bold italic M bold italic x

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

    • bold italic M equals open parentheses table row 0 1 row cell negative b end cell cell negative a end cell end table close parentheses

  • You can then investigate the solutions of the coupled differential equations

  • If the eigenvalues lambda subscript 1 and lambda subscript 2 are real, distinct, non-zero

    • The exact solution is of the form x equals C straight e to the power of lambda subscript 1 t end exponent plus D straight e to the power of lambda subscript 2 t end exponent

      • Where C and D are constants

    • You can find the value of the constants given initial or boundary conditions

      • Remember that y equals fraction numerator d x over denominator d t end fraction

      • This helps if you are given the initial value of fraction numerator d x over denominator d t end fraction

Examiner Tips and Tricks

In your exam, the eigenvalues will always be real, distinct and non-zero for these questions. The formula 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 2 is given in the formula booklet.

  • For example, consider fraction numerator d squared x over denominator d t squared end fraction minus 4 fraction numerator d x over denominator d t end fraction plus 3 x equals 0

    • The eigenvalues of open parentheses table row 0 1 row cell negative 3 end cell 4 end table close parentheses are 1 and 3

    • The general solution is x equals C straight e to the power of t plus D straight e to the power of 3 t end exponent

Worked Example

Consider the second order differential equation fraction numerator straight d squared x over denominator straight d t squared end fraction plus 3 fraction numerator straight d x over denominator straight d t end fraction minus 4 x equals 0. Initially x equals 3 and  fraction numerator d x over denominator d t end fraction equals negative 2.

a) Show that the equation above can be rewritten as a system of coupled first order differential equations.

5-7-2-ib-ai-hl-exact-second-order-a-we-solution

b) Given that the matrix open parentheses table row 0 1 row 4 cell negative 3 end cell end table close parentheses has eigenvalues of 1 and -4 with corresponding eigenvectors open parentheses table row 1 row 1 end table close parentheses and open parentheses table row cell negative 1 end cell row 4 end table close parentheses, find the exact solution to the second order differential equation.

5-7-2-ib-ai-hl-exact-second-order-b-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.

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.