Analytical Solutions to Differential Equations (DP IB Maths: AA HL)

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Roger

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Roger

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Separation of Variables

What is separation of variables?

  • Separation of variables can be used to solve certain types of first order differential equations
  • Look out for equations of the form fraction numerator d y over denominator d x end fraction equals g left parenthesis x right parenthesis h left parenthesis y right parenthesis
    • i.e. fraction numerator d y over denominator d x end fraction is a function of x multiplied by a function of y
    • be careful – the ‘function of xg left parenthesis x right parenthesis may just be a constant!
      • For example in fraction numerator d y over denominator d x end fraction equals 6 y, g left parenthesis x right parenthesis equals 6 and h left parenthesis y right parenthesis equals y
  • If the equation is in that form you can use separation of variables to try to solve it
  • If the equation is not in that form you will need to use another solution method

How do I solve a differential equation using separation of variables?

  • STEP 1: Rearrange the equation into the form open parentheses fraction numerator 1 over denominator h left parenthesis y right parenthesis end fraction close parentheses fraction numerator d y over denominator d x end fraction equals g left parenthesis x right parenthesis
  • STEP 2: Take the integral of both sides to change the equation into the form 

integral subscript blank superscript blank fraction numerator 1 over denominator h left parenthesis y right parenthesis end fraction space d y equals integral subscript blank superscript blank g left parenthesis x right parenthesis space d x

    • You can think of this step as ‘multiplying the d x across and integrating both sides’
      • Mathematically that’s not quite what is actually happening, but it will get you the right answer here!
  • STEP 3: Work out the integrals on both sides of the equation to find the general solution to the differential equation
    • Don’t forget to include a constant of integration
      • Although there are two integrals, you only need to include one constant of integration
    • Look out for integrals that require you to use partial fractions to solve them
      • See ‘Integrating with Partial Fractions’ in 5.9 Advanced Integration
  • STEP 4: Use any boundary or initial conditions in the question to work out the value of the integration constant
  • STEP 5: If necessary, rearrange the solution into the form required by the question

Examiner Tip

  • Be careful with letters – the equation on an exam may not use xand y as the variables
  • Unless the question asks for it, you don’t have to change your solution into y equals f left parenthesis x right parenthesis form – sometimes it might be more convenient to leave your solution in another form

Worked example

For each of the following differential equations, either (i) solve the equation by using separation of variables giving your answer in the form y equals f left parenthesis x right parenthesis, or (ii) state why the equation may not be solved using separation of variables.

a)       fraction numerator d y over denominator d x end fraction equals fraction numerator straight e to the power of x plus 4 x over denominator 3 y squared end fraction.

5-10-2-ib-aa-hl-separation-of-variables-a-we-solution

b)       fraction numerator d y over denominator d x end fraction equals 4 x y minus 2 ln space x.

5-10-2-ib-aa-hl-separation-of-variables-b-we-solution

c)       fraction numerator d y over denominator d x end fraction equals 2 y squared plus 2 y, given that y equals 2 when x equals 0.

5-10-2-ib-aa-hl-separation-of-variables-c-we-solution

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Homogeneous Differential Equations

What is a homogeneous first order differential equation?

  • If a first order differential equation can be written in the form fraction numerator d y over denominator d x end fraction equals f open parentheses y over x close parentheses then it is said to be homogeneous

How do I solve a homogeneous first order differential equation?

  • These equations can be solved using the substitution v equals y over x left right double arrow y equals v x
  • STEP 1: If necessary, rearrange the equation into the form fraction numerator d y over denominator d x end fraction equals f open parentheses y over x close parentheses
  • STEP 2: Replace all instances of y over x in your equation with v
  • STEP 3: Use the product rule and implicit differentiation to replace fraction numerator d y over denominator d x end fraction in your equation with v plus x fraction numerator d v over denominator d x end fraction
    • This is because y equals v x blank ⟹ blank fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight d over denominator straight d x end fraction open parentheses v x close parentheses equals v fraction numerator straight d over denominator straight d x end fraction open parentheses x close parentheses plus x fraction numerator straight d over denominator straight d x end fraction open parentheses v close parentheses equals v plus x fraction numerator straight d v over denominator straight d x end fraction
  • STEP 4: Solve your new differential equation to find the solution in terms of v and x
    • You may need to use other methods for differential equations, such as separation of variables, at this stage
  • STEP 5: Substitute v equals y over x into the solution from Step 4, in order to find the solution in terms of y and x

What else should I know about solving homogeneous first order differential equations?

  • After finding the solution in terms of y and x you may be asked to do other things with the solution
    • For example you may be asked to find the solution corresponding to certain initial or boundary conditions
    • Or you may be asked to express your answer in a particular form, such as y = f(x)
  • It is sometimes possible to solve differential equations that are not homogeneous by using the substitution v equals y over x
    • For such a situation in an exam question, you would be told explicitly to use the substitution
    • You would not be expected to know that you could use the substitution in a case where the differential equation was not homogeneous

Examiner Tip

  • Unless the question asks for it, you don’t have to change your solution into y = f(x) form – sometimes it might be more convenient to leave your solution in another form

Worked example

Consider the differential equation x y fraction numerator straight d y over denominator straight d x end fraction equals y squared minus x squared where  y = 3  when  x = 1.

a)
Show that the differential equation is homogeneous.

5-10-2-ib-aa-hl-homogeneous-diff-eqn-a-we-solution

b)
Use the substitution v equals y over x to solve the differential equation with the given boundary condition.

5-10-2-ib-aa-hl-homogeneous-diff-eqn-b-we-solution

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Integrating Factor

What is an integrating factor?

  • An integrating factor can be used to solve a differential equation that can be written in the standard form fraction numerator d y over denominator d x end fraction plus p left parenthesis x right parenthesis y equals q left parenthesis x right parenthesis
    • Be careful – the ‘functions of xp(x) and q(x) may just be constants!
      • For example in fraction numerator d y over denominator d x end fraction plus 6 y equals straight e to the power of negative 2 x end exponent, p(x) = 6 and q(x) = e-2x
      • While in fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator 2 x end fraction equals 12, p left parenthesis x right parenthesis equals fraction numerator 1 over denominator 2 x end fraction  and q(x) = 12
  • For an equation in standard form, the integrating factor is straight e to the power of   integral p open parentheses x close parentheses   straight d x end exponent

How do I use an integrating factor to solve a differential equation?

  • STEP 1: If necessary, rearrange the differential equation into standard form
  • STEP 2: Find the integrating factor
    • Note that you don’t need to include a constant of integration here when you integrate  ∫p(x) dx
  • STEP 3: Multiply both sides of the differential equation by the integrating factor
  • This will turn the equation into an exact differential equation of the form fraction numerator straight d over denominator straight d x end fraction open parentheses y straight e to the power of integral p open parentheses x close parentheses   straight d x end exponent close parentheses equals q open parentheses x close parentheses straight e to the power of   integral p open parentheses x close parentheses   straight d x end exponent
  • STEP 4: Integrate both sides of the equation with respect to x
    • The left side will automatically integrate to space y straight e to the power of integral p open parentheses x close parentheses   d x end exponent
    • For the right side, integrate integral q left parenthesis x right parenthesis straight e to the power of integral p left parenthesis x right parenthesis d x end exponent d x using your usual techniques for integration
    • Don’t forget to include a constant of integration
      • Although there are two integrals, you only need to include one constant of integration
  • STEP 5: Rearrange your solution to get it in the form y = f(x)

What else should I know about using an integrating factor to solve differential equations?

  • After finding the general solution using the steps above you may be asked to do other things with the solution
    • For example you may be asked to find the solution corresponding to certain initial or boundary conditions

Worked example

Consider the differential equation fraction numerator straight d y over denominator straight d x end fraction equals 2 x y plus 5 straight e to the power of x squared end exponent where  y = 7  when  x = 0.

Use an integrating factor to find the solution to the differential equation with the given boundary condition.

5-10-2-ib-aa-hl-integrating-factor-we-solution

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Roger

Author: Roger

Expertise: Maths

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.