Second Order Differential Equations (DP IB Maths: AI HL)

Revision Note

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Roger

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Euler's Method: Second Order

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

  • A second order differential equation is a differential equation containing one or more second derivatives
  • In this section of the course we 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

    • You may need to rearrange the differential equation given to get it in this form
  • In order to apply Euler’s method, use the substitution space y equals fraction numerator d x over denominator d t end fraction to turn the second order differential equation into a pair of coupled first order differential equations
    • If space y equals fraction numerator d x over denominator d t end fraction, then fraction numerator straight d y over denominator straight d t end fraction equals fraction numerator straight d squared x over denominator straight d t squared end fraction
    • This changes the second order differential equation into the coupled system
  • Approximate solutions to this coupled system can then be found using the standard Euler’s method for coupled systems
    • See the notes on this method in the revision note 5.6.4 Approximate Solutions to Differential Equations

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

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 = 0.5 .

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

 

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Exact Solutions & Phase Portraits: Second Order

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

  • In some cases we can apply methods we already know to find the exact solutions for second order differential equations
  • In this section of the course we consider 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

    •  are constants
  • Use the substitution space y equals fraction numerator d x over denominator d t end fraction to turn the second order differential equation into a pair of coupled first order differential equations
    • If space y equals fraction numerator d x over denominator d t end fraction, then fraction numerator straight d y over denominator straight d t end fraction equals fraction numerator straight d squared x over denominator straight d t squared end fraction
    • This changes the second order differential equation into the coupled system

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d x over denominator d t end fraction end cell equals y row cell fraction numerator d y over denominator d t end fraction end cell equals cell negative b x minus a y end cell end table

    • The coupled system may also be represented in matrix form 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 0 1 row cell negative b end cell cell negative a end cell end table close parentheses open parentheses table row x row y end table close parentheses

      • In the ‘dot notation’ here  and
    • That can be written even more succinctly as bold italic x with dot on top equals bold italic M bold italic x
      • Here 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, and 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
  • Once the original equation has been rewritten in matrix form, the standard method for finding exact solutions of systems of coupled differential equations may be used
    • The solutions will depend on the eigenvalues and eigenvectors of the matrix M
    • For the details of the solution method see the revision note 5.7.1 Coupled Differential Equations
    • Remember that exam questions will only ask for exact solutions for cases where the eigenvalues of M are real and distinct

How can I use phase portraits to investigate the solutions to second order differential equations?

  • Here we are again considering 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

    •  a & b are real constants
  • As shown above, the substitution space y equals fraction numerator d x over denominator d t end fraction can be used to convert this second order differential equation into a system of coupled first order differential equations of the form bold italic x with dot on top equals bold italic M bold italic x
    • Here 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, and 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
  • Once the equation has been rewritten in this form, you may use the standard methods to construct a phase portrait or sketch a solution trajectory for the equation
    • For the details of the phase portrait and solution trajectory methods see the revision note 5.7.1 Coupled Differential Equations
    • When interpreting a phase portrait or solution trajectory sketch, don’t forget that space y equals fraction numerator d x over denominator d t end fraction
      • So if x represents the displacement of a particle, for example, then space y equals fraction numerator d x over denominator d t end fraction will represent the particle’s velocity

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 = 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

c)
Sketch the trajectory of the solution to the equation on a phase diagram, showing the relationship between x and fraction numerator d x over denominator d t end fraction.

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

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