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Suppose a rock falls from rest from a height of 100 meters and the only force acting on it is gravity. Find an equation for the velocity v ( t ) as a function of time, measured in meters per second.

v ( t ) = −9.8 t

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A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a given point in time. Let s ( t ) denote the height above Earth’s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation s ( t ) = v ( t ) . An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation s ( 0 ) = s 0 . Together these assumptions give the initial-value problem

s ( t ) = v ( t ) , s ( 0 ) = s 0 .

If the velocity function is known, then it is possible to solve for the position function as well.

Height of a moving baseball

A baseball is thrown upward from a height of 3 meters above Earth’s surface with an initial velocity of 10 m/s , and the only force acting on it is gravity. The ball has a mass of 0.15 kilogram at Earth’s surface.

  1. Find the position s ( t ) of the baseball at time t .
  2. What is its height after 2 seconds?
  1. We already know the velocity function for this problem is v ( t ) = −9.8 t + 10 . The initial height of the baseball is 3 meters, so s 0 = 3 . Therefore the initial-value problem for this example is
    To solve the initial-value problem, we first find the antiderivatives:
    s ( t ) d t = −9.8 t + 10 d t s ( t ) = −4.9 t 2 + 10 t + C .

    Next we substitute t = 0 and solve for C :
    s ( t ) = −4.9 t 2 + 10 t + C s ( 0 ) = −4.9 ( 0 ) 2 + 10 ( 0 ) + C 3 = C .

    Therefore the position function is s ( t ) = −4.9 t 2 + 10 t + 3 .
  2. The height of the baseball after 2 s is given by s ( 2 ) :
    s ( 2 ) = −4.9 ( 2 ) 2 + 10 ( 2 ) + 3 = −4.9 ( 4 ) + 23 = 3.4.

    Therefore the baseball is 3.4 meters above Earth’s surface after 2 seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.
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Key concepts

  • A differential equation is an equation involving a function y = f ( x ) and one or more of its derivatives. A solution is a function y = f ( x ) that satisfies the differential equation when f and its derivatives are substituted into the equation.
  • The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.
  • A differential equation coupled with an initial value is called an initial-value problem. To solve an initial-value problem, first find the general solution to the differential equation, then determine the value of the constant. Initial-value problems have many applications in science and engineering.

Determine the order of the following differential equations.

( y ) 2 = y + 2 y

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y + y y = 3 x 2

3

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d y d x + d 2 y d x 2 = 3 x 4

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( d y d t ) 2 + 8 d y d t + 3 y = 4 t

1

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Verify that the following functions are solutions to the given differential equation.

y = x 3 3 solves y = x 2

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y = 2 e x + x 1 solves y = x y

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y = e 3 x e x 2 solves y = 3 y + e x

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y = 1 1 x solves y = y 2

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y = e x 2 / 2 solves y = x y

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y = 4 + ln x solves x y = 1

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y = 3 x + x ln x solves y = ln x

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y = 2 e x x 1 solves y = y + x

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y = e x + sin x 2 cos x 2 solves y = cos x + y

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y = π e cos x solves y = y sin x

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Verify the following general solutions and find the particular solution.

Practice Key Terms 8

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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