9780070109100, Chapter 15 5, Problem 1E

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Chapter 15.5, Problem 1E Step-by-step solution Step 1 of 4 1. (a) i. Separable; we can write the equation as + 4dt = 0. ii. Rewritten as the equation is linear. Step 2 of 4 (b) i. Separable; multiplying by (y + t), we get y dy + 2t dt = 0. ii. Rewritten as = the equation is a Bernoulli equation with R = 0, T = -2t and Define Z = y1-m = y2. Then we can obtain from (15.24') a linearized equation dz 2(-2)t dt = 0 or Step 3 of 4 (c) i. Separable; we can write the equation as y dy +t dt = 0. ii. Reducible; it is a Bernoulli equation with and m = -1. Step 4 of 4 (d) i. Separable; we can write the equation as t dt = 0 ii. Yes; it is a Bernoulli equation with T = 3t, m = 2.