AP®MathsCollege BoardCalculus ABRevision NotesUnit 7: Differential EquationsFirst-Order Differential EquationsFinding Particular Solutions

Finding Particular Solutions (College Board AP® Calculus AB)

Revision Note

Roger B

Author

Roger B

Expertise

Maths

Initial conditions

How can I use initial conditions to find the particular solution of a differential equation?

  • Remember that the general solution of a differential equation represents a family of solutions

    • In general there will be an infinite number of possible solutions

    • Each of these possible solutions is known as a particular solution

  • You need additional information to determine a particular solution

    • This additional information is known as an initial condition (or sometimes boundary condition)

    • For example, you might be given a value of y that corresponds to a particular value of x

  • If you think of the family of solutions as a family of curves on a graph

    • then there is only one particular solution that passes through a given point

      • So if you know a point the solution curve goes through

      • then you can determine the unique particular solution

  • Consider the simple case where fraction numerator d y over denominator d x end fraction equals f open parentheses x close parentheses

    • The general solution can be found by differentiation: space y equals integral f open parentheses x close parentheses space d x

      • This solution will contain a constant of integration, and so represent an infinite number of possible solutions

    • But if you know that the solution goes through the point open parentheses a comma space y subscript 0 close parentheses, then the particular solution to the equation is

      • F open parentheses x close parentheses equals y subscript 0 plus integral subscript a superscript x f open parentheses t close parentheses space d t

        • Note that F open parentheses a close parentheses equals y subscript 0 plus integral subscript a superscript a f open parentheses t close parentheses space d t equals y subscript 0 plus 0 equals y subscript 0, as required

    • Note as well that this is an alternative approach to 'finding the constant of integration'

Worked Example

The function F defined by F open parentheses x close parentheses equals y subscript 0 plus integral subscript a superscript x f open parentheses t close parentheses space d t is a particular solution to the differential equation fraction numerator d y over denominator d x end fraction equals f open parentheses x close parentheses, satisfying F open parentheses a close parentheses equals y subscript 0.

Use this fact to find the particular solution to the differential equation fraction numerator d y over denominator d x end fraction equals 3 over x, given that y equals 1 when x equals e.

Answer:

Here y subscript 0 equals 1 and a equals e

Recall that ln open parentheses e close parentheses equals 1

table row cell F open parentheses x close parentheses end cell equals cell 1 plus integral subscript e superscript x 3 over t space d t end cell row blank equals cell 1 plus 3 integral subscript e superscript x 1 over t space d t end cell row blank equals cell 1 plus 3 open square brackets ln open vertical bar t close vertical bar close square brackets subscript e superscript x end cell row blank equals cell 1 plus 3 open parentheses ln open vertical bar x close vertical bar minus ln open vertical bar e close vertical bar close parentheses end cell row blank equals cell 1 plus 3 open parentheses ln open vertical bar x close vertical bar minus 1 close parentheses end cell row blank equals cell 3 ln open vertical bar x close vertical bar minus 2 end cell end table

y equals 3 ln open vertical bar x close vertical bar minus 2

Finding particular solutions using separation of variables

  • If you are given an initial condition, then you can find the particular solution to a differential equation solved by separation of variables:

    • Substitute the initial condition values into the general solution

    • and solve to find the value of the arbitrary constant

  • E.g. the general solution to space fraction numerator d y over denominator d x end fraction equals negative x y to the power of 3 space end exponentis space y squared equals fraction numerator 1 over denominator x squared plus C end fraction

    • If you know that y equals 1 third when x equals 2, then

      • space open parentheses 1 third close parentheses squared equals fraction numerator 1 over denominator open parentheses 2 close parentheses squared plus C end fraction space space rightwards double arrow space space 1 over 9 equals fraction numerator 1 over denominator 4 plus C end fraction space space rightwards double arrow space space C plus 4 equals 9 space space rightwards double arrow space space C equals 5

    • So the particular solution satisfying that initial condition is

      • y squared equals fraction numerator 1 over denominator x squared plus 5 end fraction

Worked Example

Find the solution to the differential equation space open parentheses x plus 3 close parentheses fraction numerator d y over denominator d x end fraction equals sec space y, given that the graph of the solution goes through the point open parentheses negative 2 comma fraction numerator space 3 pi over denominator 2 end fraction close parentheses .

Answer:

Separate the variables, getting all the y terms on the fraction numerator d y over denominator d x end fraction side and all the x terms on the other side

Recall that sec y equals fraction numerator 1 over denominator cos y end fraction

open parentheses x plus 3 close parentheses fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator cos y end fraction cos y fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator x plus 3 end fraction

Integrate both sides with respect to x

integral cos y space d y equals integral fraction numerator 1 over denominator x plus 3 end fraction space d x

Integrate (and don't forget a constant of integration)

sin y equals ln open vertical bar x plus 3 close vertical bar plus C

Don't try to rewrite that as y equals arcsin open parentheses ln open vertical bar x plus 3 close vertical bar plus C close parentheses; that would lose parts of the solution

Now bring in the boundary condition, y equals fraction numerator 3 pi over denominator 2 end fraction when x equals negative 2

Recall that ln open parentheses 1 close parentheses equals 0

table row cell sin open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses end cell equals cell ln open vertical bar open parentheses negative 2 close parentheses plus 3 close vertical bar plus C end cell row cell negative 1 end cell equals cell 0 plus C end cell row C equals cell negative 1 end cell end table

Substitute that value of C into the general solution

sin y equals ln open vertical bar x plus 3 close vertical bar minus 1

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    Roger B

    Author: Roger B

    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.