By Erwin Kreyszig
Ebook by way of Kreyszig, Erwin
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Team Project. (A) We obtain ( Ϫ 1)( Ϫ 2) ϭ 2 Ϫ (1 ϩ 2) ϩ 12 ϭ 2 ϩ a ϩ b ϭ 0. Comparison of coefficients gives a ϭ Ϫ(1 ϩ 2), b ϭ 12. (B) y Љ ϩ ayЈ ϭ 0. (i) y ϭ c1e؊ax ϩ c2 e 0x ϭ c1e؊ax ϩ c2. (ii) zЈ ϩ az ϭ 0, where z ϭ yЈ, z ϭ ce؊ax and the second term comes in by integration: y ϭ ͐ z dx ϭ ෂc1e؊ax ϩ ෂc2. (D) e(kϩm)x and ekx satisfy y Љ Ϫ (2k ϩ m)yЈ ϩ k(k ϩ m)y ϭ 0, by the coefficient formulas in part (A). By the superposition principle, another solution is e(kϩm)x Ϫ ekx ᎏᎏ . m We now let m * 0.
Hence a general solution of the homogeneous ODE is e؊t(A cos 2t ϩ B sin 2t). 5t. 16. 1I Љ ϩ 4IЈ ϩ 40I ϭ 100 cos 10t. 6 sin 10t. The initial conditions are I(0) ϭ 0, Q(0) ϭ 0, which because of (1Ј), that is, Q(0) LIЈ(0) ϩ RI(0) ϩ ᎏ ϭ E(0) ϭ 0, C leads to IЈ(0) ϭ 0. 2 IЈ(0) ϭ Ϫ20c1 ϩ c2 ϩ 16 ϭ 0, c2 ϭ Ϫ40. 6 sin 10t. 18. 2( ϩ 8)( ϩ 10) ϭ 0. 2 ϭ 820 by formula (1Ј) in the text and Q(0) ϭ 0. Also, EЈ ϭ Ϫ1640 sin 10t. 6IЈ ϩ 16I ϭ Ϫ1640 sin 10t. The answer is I ϭ 160 e؊8t Ϫ 205e؊10t ϩ 45 cos 10t ϩ 5 sin 10t.
Formula (4) in Sec. 5 gives, since p ϭ Ϫ3, h ϭ Ϫ3x, ͵ y ϭ e 3x ( e؊3x(Ϫ2x) dx ϩ c) ϭ e 3x(e؊3x(_32x ϩ _92 ) ϩ c) ϭ ce 3x ϩ _32x ϩ _92. 14. Separate variables. y dy ϭ 16x dx, _12y 2 ϭ 8x 2 ϩ c*, y 2 Ϫ 16x 2 ϭ c. Hyperbolas. 16. Linear ODE. Standard form yЈ Ϫ xy ϭ Ϫx 3 ϩ x. Use (4), Sec. 5, with p ϭ Ϫx, h ϭ Ϫx 2/2, obtaining the general solution ͵ y ϭ ex /2 ( e؊x /2 (Ϫx 3 ϩ x) dx ϩ c) ϭ ex /2[e؊x /2(x 2 ϩ 1) ϩ c] 2 2 2 2 2 ϭ cex /2 ϩ x 2 ϩ 1. 18. Exact; the exactness test gives Ϫ3 sin x sinh 3y on both sides.