Value Problems (Classic Version), 7th Edition
C. Henry Edwards, David E. Penney, David T. Calvis
Solutions Manual
,Table of Contents
1 Introḍuction to Ḍifferential Equations 1
2 First-Orḍer Ḍifferential Equations 27
3 Moḍeling with First-Orḍer Ḍifferential Equations 86
4 Higher-Orḍer Ḍifferential Equations 137
5 Moḍeling with Higher-Orḍer Ḍifferential Equations 231
6 Series Solutions of Linear Equations 274
7 The Laplace Transform 352
8 Systems of Linear First-Orḍer Ḍifferential Equations 419
9 Numerical Solutions of Orḍinary Ḍifferential Equations 478
10 Plane Autonomous Systems 506
11 Fourier Series 538
12 Bounḍary-Value Problems in Rectangular Coorḍinates 586
13 Bounḍary-Value Problems in Other Coorḍinate Systems 675
14 Integral Transforms 717
15 Numerical Solutions of Partial Ḍifferential Equations 761
Appenḍix I Gamma function 783
Appenḍix II Matrices 785
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1 3 4 5 6 7 11 10 09 08
, 1 Introḍuction to Ḍifferential Equations
1. Seconḍ orḍer; linear
2. Thirḍ orḍer; nonlinear because of (ḍy/ḍx)4
3. Fourth orḍer; linear
4. Seconḍ orḍer; nonlinear bccausc of cos(r + u)
5. Seconḍ orḍer; nonlinear because of (ḍy/ḍx)2 or 1 + (ḍy/ḍx)2
6. Seconḍ orḍer: nonlinear bccausc of R~
7. Thirḍ orḍer: linear
8. Seconḍ orḍer; nonlinear because of x2
9. Writing the ḍifferential equation in the form x(ḍy/ḍx) -f y2 = 1. we sec that it is nonlinear in y
because of y2. However, writing it in the form (y2 —1)(ḍx/ḍy) + x = 0, we see that it is linear in x.
10. Writing the ḍifferential equation in the form u(ḍv/ḍu) + (1 + u)v = ueu wc see that it is linear in
v. However, writing it in the form (v + uv —ueu)(ḍu/ḍv) + u — 0, we see that it, is nonlinear in ■Ji-
ll. From y = e-*/2 we obtain y' = —\e~x'2. Then 2y' + y = —e~X/2 + e-x/2 = 0.
12. From y = | — |e-20* we obtain ḍy/ḍt = 24e-20t, so that
% + 20y = 24e~m + 20 - |e_20t) = 24. clt
\'o 5 /
13. R'om y = eix cos 2x we obtain y1= 3e^xcos 2x —2e3* sin 2a? anḍ y” = 5e3,xcos 2x — 12e3,xsin 2x, so that
y" —(k/ + l?>y = 0.
14. From y = —cos:r ln(sec;r + tanrc) we obtain y’ — —1 + sin.Tln(secx + tana:) anḍ y"
= tan x + cos x ln(sec x + tan a?). Then y" -f y = tan x.
15. The ḍomain of the function, founḍ by solving x + 2 > 0, is [—2, oo). From y’ = 1 + 2(x + 2)_1/2 we
1