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Example text
1 Direction Fields First, let us examine the geometric significance of the differential equation (1) y = f(x, y). At each point (x, y) in the xy-plane for which the function f is defined, the differential equation defines a real value, f(x, y). This value is the slope of the tangent line to every solution of the differential equation which passes through the point (x, y). Thus, the differential equation specifies the direction that a solution must have at every point (x, y) in the domain of f. Imagine passing a short line segment of slope f(x, y) through each point (x, y) in the domain of f.
Imposing the initial condition y(a) = b, we find c must satisfy the equation y(a) = b = ca. Since we have assumed a = 0, the unique solution of this equation is c = b/a; and, therefore, the unique solution of the initial value problem xy − y = 0; y(a) = b where a = 0 is y(x) = bx/a. Interpreted geometrically, this means that for any point (a, b) where a = 0—that is, for any point which is not on the y-axis—there is a unique solution of the differential equation xy − y = 0 which passes through (a, b).
Hence, there is no solution to the boundary value problem (16). EXAMPLE 8 A Boundary Value Problem with a Unique Solution Now consider the boundary value problem (17) y + y = 0; y(0) = 0, y(π/2) = 1. Again, if a solution of this boundary value problem exists, it must be of the form y = A sin x. Imposing the second boundary condition, y(π/2) = 1, yields y(π/2) = 1 = A sin(π/2) = A. So A = 1 and the unique solution of the boundary value problem (17) is y = sin x. © 2010 by Taylor and Francis Group, LLC Introduction 27 EXAMPLE 9 A Boundary Value Problem with an Infinite Number of Solutions As in the previous two examples, if the boundary value problem y + y = 0; y(0) = 0, y(π) = 0 (18) is to have a solution, it must be of the form y = A sin x.
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