newell_lec11-15_100713

# I vectors writing any function which is piecewise

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Unformatted text preview: refore the supposition that W (y1 , y2 ) = 0 at some point on I is false and W (y1 , y2 ) ￿= 0 for points on I . We now know what we mean by essentially diﬀerent. Let y1 (t), y2 (t) be two l.i. solutions of (6.15). Then the linear combination y (t) = Ay1 (t) + By2 (t) satisﬁes ￿ (6.15) and can satisfy the initial conditions y (t0 ) = y0 , dy (t0 ) = y0 for all t0 ∈ I dt ￿ and all choices of real numbers y0 , y0 . Let us summarize: For this course, our main interest is in identifying linearly independent solutions y1 (t), y2 (t) of (6.15) with coeﬃcients p(t), q (t) continuous on the interval I of interest. For this purpose, the linear independence of y1 (t), y2 (t) on I is equivalent to the non-vanishing of the Wronskian on I . The equivalence is broken if our interval includes points at which p(t) and/or q (t) are discontinuous. Example: Take f (t) = t, g (t) = t2 on the interval I = (−1, 1). Then t and t2 are certainly l.i. on (−1, 1). [Since k1 t + k2 t2 = 0 for all t ∈ (−1, 1) implies that k1 = k2 = 0. Put for example t = 1/2 or t = 1/4.] They are also solutions to 2 2 y ￿￿ y ￿ + 2 y = 0. t t But the Wronskian, W (f, g ) = t2 , vanishes at t = 0. Note that p, q are discontinuous ￿t 2 at t = 0 and exp t0 t￿ dt￿ is not deﬁned if t and t0 have opposite signs. Example: Take f (t) = t3 , g (t) = t2 |t| on the interval I = (−1, 1). Then W ≡ 0 for all t ∈ I yet t3 and t2 |t| are l.i. on I . [Take k1 t3 + k2 t2 |t| = 0 for 20 all t ∈ (−1, 1). Put t = 1/2, −1/2 ⇒ k1 = k2 = 0.] Also t3 and t2 |t| are solutions to 3 3 y ￿￿ − y ￿ + 2 y = 0 t t ￿t 3 But again p, q are discontinuous at t = 0 and exp t0 t￿ dt￿ is not deﬁned if t and t0 have opposite signs. In many other contexts (e.g. writing any vector in R3 as a linear combination of three l.i. vectors; writing any function which is piecewise continuous on (−L, L) as π π a linear combination of the l.i. basis ‘vectors’, {cos nLx , sin nLx }∞ , we will need n=0 the notion of linear independence. The ‘Wronskian’ will not always be the most useful way to test. 21 Homework Lectures 11-15 Question 1: For each of the following equations, give the (a) order and say whether each is (b) linear or nonlinear (c) autonomous or nonautonomous (i) dy dt + 3ty = 5 dy dx (ii) + y 2 = 3y + 10 (iii) d3 y dt3 + y dy + cos t = 0 dt (iv) d2 y dt2 + 0.2 dy + y = cos ω t dt (v) d2 y dx2 (vi) dy dx (vii) + xy = 0 = −x y d4 y dt4 =0 Question 2: Solve the following initial value problems: (i) dy dt (ii) + y = e−t , dy dt = y (1 − y ), (iii) d2 y dt2 (iv) dy dx (v) dy dx y (0) = 1 (a) y (0) = 2; (b) y (0) = 1 . 2 + 2 dy + 5y = cos 2t, dt = −x, y √ = + y, y (0) = dy dt (0) = 0. y (1) = 1. y (0) = 0. Is there more than one solution? 23 Question 3: A tank initially contains 100 gallons of brine with a concentration of 1/2 lb. per gallon. At time t = 0, an inlet valve is turned on which carries a new brine solution into the tank at a rate of 2 gallons per mi...
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## This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.

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