Introduction to Differential Equations (College Board AP® Calculus AB): Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Modeling with differential equations

What is a differential equation?

  • A differential equation is simply an equation that contains derivatives

    • For example  dydx=12xy2  is a differential equation

    • And so is  d2xdt25dxdt+7x=5sint

  • It is an equation that includes both variables and rates of change of those variables

What is a first order differential equation?

  • A first order differential equation is a differential equation that contains first derivatives but no second (or higher) derivatives

    • For example  dydx=12xy2  is a first order differential equation

    • But  d2xdt25dxdt+7x=5sint  is not a first order differential equation

      • because it contains the second derivative d2xdt2

Why are differential equations useful for modeling?

  • Many quantities of interest in the real world involve rates of change

    • For example:

      • The rate of change of a population of animals in a geographic area

      • The rate of change of the number of people infected by a particular disease

      • The rate of change of the amount of medication in a person's bloodstream at different times after ingestion

      • The rate of change of velocity for a falling object

      • The rate of change of voltage across a component in an electrical circuit

  • If the relationship between quantities and their rates of change can be written as a differential equation

    • then solving the equation can allow us to predict the behavior of the quantities in the real world

General and particular solutions to differential equations

What is the difference between general and particular solutions for a differential equation?

  • The general solution to a differential equation may be thought of as 'every possible solution' to the differential equation

    • E.g.  y2=Cx24 is the general solution to dydx=x4y

      • C is an arbitrary constant (like a constant of integration)

    • The general solution is actually an infinite family of solutions

      • Each one corresponding to a different value of C

    • The graph of the solution will change depending on the value of C

      • This is shown in the diagram below:

A graph shows nested ellipses corresponding to the equation y² = (C - x²)/4 with different values of C (1, 4, 9), centered at the origin, with labeled x and y axes.
  • The particular solution to a differential equation is

    • the specific member of the general family of solutions

      • that satisfies the equation under a particular set of conditions

    • E.g. if we know that y=1 when x=0

      • then  y2=4x24 is the only solution that satisfies the differential equation with that set of conditions

    • A condition like "y=1 when x=0" is known as an initial condition (or boundary condition)

      • Finding a particular solution requires knowing an initial condition

Verifying solutions to differential equations

How can I use differentiation to verify solutions for a differential equation?

  • You can differentiate an answer to a differential equation to verify that it is indeed a solution

    • E.g. verify that y=1xcosx is a solution to the differential equation dydx=sinx1

      • Differentiate the proposed answer with respect to x

        • dydx=ddx(1xcosx)=1+sinx=sinx1

      • This matches the original differential equation

        • So the solution has been verified

  • For more complicated answers this may require the use of additional techniques

    • E.g. implicit differentiation and/or substitution

    • See the Worked Example

Worked Example

Verify that  y2=1x2+C, where C is an arbitrary constant, is a solution to the differential equation  dydx=xy3.

Answer:

Start by differentiating both sides of the proposed solution with respect to x, using implicit differentiation

 y2=(x2+C)1ddx(y2)=ddx((x2+C)1)2ydydx=(x2+C)2·2x2ydydx=2x(x2+C)2

Rearrange to make dydx the subject

dydx=xy(x2+C)2

This may not look like  dydx=xy3

But remember that  y2=1x2+C

dydx=xy·1(x2+C)2=xy·(1x2+C)2=xy·(y2)2=xy·y4=xy3

This confirms that  y2=1x2+C is a solution

 y2=1x2+C is a solution to the differential equation  dydx=xy3

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Roger B

Author: Roger B

Expertise: Development Editor

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: Portfolio 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.