homework 01 solutions

# homework 01 solutions - Section 1.1 26 Substitution of x =...

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Unformatted text preview: Section 1 .1 26. Substitution of x = 7t and y = 0 into y = (x+ C) cosx yields the equation 0 = (75+C)(—1), so C = —7r. 10 28. The slope of the line through (x, y) and (x/2,0) is y’ = (y — 0) /(x — x/2) = 2y Ix, so the differential equation is x y’ = 2 y. 34. dv/dt = k(250-v) 41. We reason that if y = ke", then each term in the differential equation is a multiple of e". The choice k = % balances the equation, and provides the solution y(x) = %e". Section 1.2 16. If a(t) =1/Jt+4 then v(t) = I1/Jt+4 dt = 2Jt+4+C = 2Jt+4—5 (taking C=—5 so that v(0)=—1). Hence x(t) = j(2Jt+4—5)dt= %(t+4)3/2—5t+C = %(t+4)3'2—5t—% 36. The method of solution is precisely the same as that in Problem 30. We ﬁnd ﬁrst that, on Earth, the woman must jump straight upward with initial ve10city v0 = 12 ft/sec to reach a maximum height of 2.25 ft. Then we ﬁnd that, on the Moon, this initial velocity yields a maximum height of about 13.58 ft. 42. Let x(t) be the (positive) altitude (in miles) of the spacecraft at time t (hours), with t = Ocorresponding to the time at which the its retrorockets are fired; let v(t) = x’(t) be the velocity of the spacecraft at time t. Then v0 = —1000 and x0 = x(0) is unknown. But the (constant) acceleration is a = +20000, so v(t) = 20000t—1000 and x(t)=10000t2-1000t+xo. Now v(t) = 20000t— 1000 = 0 (soft touchdown) when t = 315 hr (that is, after exactly 3 minutes of descent. Finally, the condition 0 = x(—2‘5)=10000(% 2—1000(%)+x0 yields x0 = 25 miles for the altitude at which the retrorockets should be fired. Section 1 .3 15. 22. f (x, y) = (x — y)“2 is not continuous at (2, 2) because it is not even deﬁned if y > x. Hence the theorem guarantees neither existence nor uniqueness in any neighborhood of the point x = 2. Tracing the curve in the figure , we see that y(——4) z -3. An exact solution of the differential equation yields the more accurate approximation y(—4) = -3.0017. 32. ‘ Looking at the ﬁgure , we see that we can piece together a "left half" of a quartic for x negative, an interval along the x—axis, and a "right half" of a quartic curve for x positive. This makes it clear he initial value problem y' = 4x\/—, y(a) = b has inﬁnitely many solutions (deﬁned for all x) if b 2 0; there is no solution if b < 0 because this would involve the square root of a negative number. ...
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