F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester's A Modern Introduction to Differential Equations PDF

By F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester

ISBN-10: 1852338962

ISBN-13: 9781852338961

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And y (n−1) (t0 ) = yn−1 , where y0 , y1 , . . , yn−1 are arbitrary real constants, is called solving an initial-value problem (IVP). The n specified values y(t0 ) = y0 , y (t0 ) = y1 , y (t0 ) = y2 , . . , and y (n−1) (t0 ) = yn−1 are called initial conditions. Right now we can’t be sure of the circumstances under which we can solve such an initial-value problem. 8. 9 we will consider IVPs for systems of differential equations. Boundary-Value Problems For second- and higher-order differential equations, we can also determine a particular solution by specifying what are called boundary conditions.

The next two examples should clarify the difference between linear and nonlinear operators. 1 A Linear Operator We can check that y1 = e−x is a solution of the homogeneous linear equation L(y) = y + y = 0 = Q1 and that y2 = sin x is a solution of L(y) = y + y = cos x + sin x = Q2 . ) You should see that y1 + y2 = e−x + sin x is a solution of the equation y + y = Q1 + Q2 = 0 + cos x + sin x = cos x + sin x—that is, ■ that L(y1 + y2 ) = Q1 + Q2 = L(y1 ) + L(y2 ). However, not every operator defined by a first-order equation is linear.

Show that the first-order nonlinear equation (xy − y)2 − (y )2 − 1 = 0 has a one-parameter family of solutions given by y = Cx ± C 2 + 1, but that any function y defined implicitly by the relation x 2 + y 2 = 1 is also a solution—one that does not correspond to a particular value of C in the one-parameter solution formula. 10. Find a differential equation satisfied by the function y(t) = cos t + t 0 (t − u)y(u)du. C 1. Consider the equation xy − (x + n)y + ny = 0, where n is a nonnegative integer.

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A Modern Introduction to Differential Equations by F.M. Dekking, C. Kraaikamp, H.P. Lopuhaä, L.E. Meester


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