A Second Course in Elementary Differential Equations - download pdf or read online

By Paul Waltman, Mathematics

ISBN-10: 0127339108

ISBN-13: 9780127339108

Focusing on appropriate instead of utilized arithmetic, this flexible textual content is acceptable for complicated undergraduates majoring in any self-discipline. a radical exam of linear structures of differential equations inaugurates the textual content, reviewing ideas from linear algebra and easy conception. the center of the publication develops the tips of balance and qualitative habit. beginning with two-dimensional linear platforms, the writer experiences using polar coordinate options in addition to Liapunov balance and effortless rules from dynamic structures. life and forte theorems obtain a rigorous remedy, and the textual content concludes with a survey of linear boundary price difficulties. 1986 ed. 39 figures.

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Extra info for A Second Course in Elementary Differential Equations

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We illustrate the procedure with some examples. Consider the system det(^ - λΐ) = deti " ~ j = 0 yields λ2 4- 1 = 0 or λ = ±L Fix λ = i. To find an eigenvector, it is necessary to solve 6. THE CONSTANT COEFRCIENT CASE 43 or — icl — c2 = 0 c i — * c2 = 0. Then c1 = 1, c2 = — i is a nontrivial solution to this linear system of equations, so a solution vector is given by Making use of Euler's formula, ew = cos(0) + isin(0), ç»(0 = (cos(0 + isin(;)( = /cos(f)\ Vsin(i)/ 7 sin(i)\ V-cos(i)/' Thus, two solutions are given by Re

3) 42 CHAPTER 1 / SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS Im [eia+iß)t(a + ib)] = cei[(sin (ßt))a + (cos (ßt))b\. 4) It is not difficult to show that these two vectors are linearly independent. At first glance it would seem that from one solution, two linearly indepen­ dent vectors have been created. This is not the case, and we explore this point in somewhat more detail. First of all, since the matrix A has real coefficients, det (A — λΐ) = ρ(λ) is a polynomial with real coefficients. Let p(X) = λη + a^"'1 + ••· + αη.

We illustrate the Putzer algorithm with a simple example. o 1 -k A 0 0 0 0 , -X -1 Expanding the determinant yields σ 0 ,1=0. 1 -K 7. THE CONSTANT COEFFICIENT CASE: THE PUTZER ALGORITHM 57 ρ(λ) = λ2(λ2 + 1) + (λ2 + 1) = (λ2 + l) 2 = 0 or that λ= ±i are double roots. To apply the algorithm, label Λ·ι = h K = -U λ3 = i, /l 4 = - i . First of all, •■e- i 1 0 0 1 —I °\ 0 0 0 —I 0 -1 °1 /l -ij P2 = (A + U)Pi = 0 (the null matrix), and hence p3 = o. Also, rx(t) satisfies the equation r[ = irl9 r1(0)=l, so rx(t) = e*\ and r2(t) satisfies r'2 = —ir2 H- eu.

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A Second Course in Elementary Differential Equations by Paul Waltman, Mathematics


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