By Dennis G. Zill

**Read or Download Complete Solution Manual - A First Course in Differential Equations with Modeling Applications 9th, Differential Equations with Boundary-Value Problems 7th PDF**

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**Extra resources for Complete Solution Manual - A First Course in Differential Equations with Modeling Applications 9th, Differential Equations with Boundary-Value Problems 7th**

**Example text**

When 0 < y < 1 we have 2 2 V1-y lny - | ln (l - y) - ^ ln(l. + y) = In ■ ■, —.. = :c + c. Letting x = 0 and y = | we find c = ln(l/\/3). Solving for y we get yi{x) — ex/y/e2x + 3, where —OO < X < oo. “ ( ! - » ) - 5 l" (l + V) = to VT-y , V = x + c. Letting x = 0 and y — —\we find c = ln(l/\/3). Solving for y we get ys(x) = —ex/sj~e2x + 3 , where —oc < x < oo. When y < —1 wc have H ~ y ) - \M 1 - y ) - \ M - 1 - v) = ln 2 2 , „ 43 = x + c. 2 Separable Variables Letting rr = 0 and y — —2 we find c = ln(2/\/3).

R and u = --- - e x ---- - e x for —1 < x < 00. The entire solution is transient. x + 2)2 and y - ^(a; + 2)_1 + c(x + 2)~4 for —2 < x < o o . The entire solution is transient, o dv For — + r see 6 dt) — cos 9 an integrating factor is eJ s<]c()d0 = ein |seen-tan_ sec q _|_ tan /9 so that ~ [(sec 9 + tan 6)r] — 1 + sin 9 and (sec 9 + tan 9)r = 9 — cos 9 + c for —tt/2 < 9 < 7t/2 . id + (21 — 1)P — 4t — 2 an integrating factor is g/(2t-1)rff = so that dt dt ■it —2)et and P = 2 + cei_ r for —oc < t < 00.

Using implicit differentiation we get yf = (x — x2)/(y2 — 2y), which is infinite when y — 0 and y = 2. 13232. 71299. 71299). (c) The value of c corresponding to y(0) = —2 is /(0, —2) = y —40. The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find dy/dx for 21/3 — 6y2 + 2xi — 3a:2 = —40. Using implicit differentiation we get y' = (x — x2)/(y2 — 2y), which is infinite when y = 0 and y = 2.