Module 20  Differential Equations and Euler's Method  
Introduction  Lesson 1  Lesson 2  Lesson 3  SelfTest  
Lesson 20.3: SecondOrder Differential Equations  
In Module 18 you learned how to use the deSolve command and the TI89 Differential Equation graphing mode to solve firstorder differential equations. In this lesson you will learn to use these tools to solve secondorder differential equations. Defining a SecondOrder Differential Equation A secondorder differential equation contains a second derivative. If y is the height above ground of a falling object then the firstorder derivative y' is the object's velocity and the second derivative y" is its acceleration. The equation y" = k is a secondorder differential equation that represents the movement of an object that has constant acceleration k. Initialvalue problems that involve a secondorder differential equation have two initial conditions. Solving SecondOrder Differential Equations Suppose that an object is dropped from a height of 48 feet. Because it is dropped without thrust, the initial velocity is zero. The only force acting on the object is gravity, which accelerates it downward 32 feet per second each second. The height of the object is given by the solution to the secondorder differential equation y" = – 32 with initial conditions y(0) = 48 feet and y'(0) = 0 feet per second. Solve the secondorder differential equation to find the equation for the height of the object.
Enter '' by pressing twice. The height of the object is given by y = 48 – 16 t^{2} . Determine the object's height above the ground at t = 1.3 seconds after it was dropped.
The height of the object at t = 1.3 is 20.96 feet.
20.3.1 Find the height of the object at t = 1.7 seconds. Only firstorder differential equations can be entered in the Y= Editor in Differential Equation graphing mode. However, by making a few substitutions you can still use that mode to solve a secondorder differential equation. Let y1 represent the height of an object. If we let y_{2} = y_{1}', then y_{2} represents the velocity of the object and y_{2}' = y_{1}" represents its acceleration. The initial height is 48 ft and the initial velocity is 0 ft/s.
20.3.2 One of the graphs is height and the other is velocity. Which is which?
20.3.3 Approximate the velocity of the object just before it hits the ground by tracing on the height curve to the xintercept then pressing
to jump to the velocity graph.


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