MAT137Y1 Lecture : MAT 137Y 2007-08Winter Session, Self Generated Solutions to Problem Set 11.pdf
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Mat 137y 2007-08 winter session, solutions to problem set 10. 1 (she 7. 4: y = x2ex xex2 = y(cid:48) = 2xex + x2ex ex2 2x2ex2. 30. (cid:90) xe x2 dx = 1 (cid:90) sin(e 2x) 40: if y = ex then x = lny. The area of the rectangle is a = ylny. Therefore e2x dx = e 2x sin(e 2x)dx = cos(e 2x) +c. At y = 3, we have dy dt = 1. 3 ln4: (she 7. 6, suppose f (cid:48)(t) = (sint) f (t). Then f (cid:48)(t) (sint) f (t) = 0. Hence f (t) = ce cost. as f (cid:48)(t) g(t) f (t) = 0 and set h(t) = (cid:82) g(t) dt. Then eh(t) f (cid:48)(t) g(t)eh(t) f (t) = 0 = (cid:104) eh(t) f (t) (cid:105)(cid:48) Therefore f (t) = ce h(t) = ce (cid:82) g(t) dt. = 0 = eh(t) f (t) = c: (she 7. 7)
