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SSM-Ch01 - Chap 1 Sec 1.1 First-Order ODEs Basic Concepts...

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Chap . 1 First - Order ODEs Sec . 1 . 1 Basic Concepts . Modeling Problem Set 1 . 1 . Page 8 1 . Calculus . To solve y u2032 u003d f ue0a2 x ue0a3 u003d u2212 sin u03c0 x is a problem of calculus, namely, to integrate u2212 sin u03c0 x , obtaining (chain rule!) y u003d 1 u03c0 cos u03c0 x u002b c . The constant of integration c is arbitrary. This is essential. It means that the ODE y u2032 u003d u2212 sin u03c0 x has infinitely many solutions, each of these congruent cosine curves corresponding to a certain value of c . For better understanding sketch some of them, for instance, those for which c u003d 0, u2212 1,1,1/2. 13 . Initial value problem ( IVP ). Differentiation of y u003d ce u2212 x 2 gives y u2032 u003d u2212 2 cxe u2212 x 2 . Hence the left side of the ODE becomes y u2032 u002b 2 xy u003d u2212 2 cxe u2212 x 2 u002b 2 x ue0a2 ce u2212 x 2 ue0a3 u003d 0. This verifies that y u003d ce u2212 x 2 is a solution of the given ODE. For x u003d 1 it has the value y ue0a2 1 ue0a3 u003d ce u2212 1 . This should equal the given initial value y ue0a2 1 ue0a3 u003d 1/ e . Hence c u003d 1, and the answer (the solution of the IVP) is y u003d e u2212 x 2 . 17 . Modeling : Falling body . y u2032u2032 u003d g u003d const is the model of the problem, an ODE of second order. Integrate to get the velocity v u003d y u2032 u003d gt u002b c 1 ( c 1 arbitrary). Integrate once more to obtain the distance fallen y u003d 1 2 gt 2 u002b c 1 t u002b c 2 ( c 2 arbitrary). All this is calculus. From this we obtain y u003d 1 2 gt 2 by imposing the initial conditions y ue0a2 0 ue0a3 u003d c 2 u003d 0 and v ue0a2 0 ue0a3 u003d y u2032 ue0a2 0 ue0a3 u003d c 1 u003d 0, meaning that the stone should start from the initial position y u003d 0 with initial velocity 0. 19 . Airplane takeoff . The solution is given on p. A4. The acceleration is y u2032u2032 u003d k with unknown k . Two integrations give the speed v u003d y u2032 u003d kt u002b c 1 and the distance y u003d 1 2 kt 2 u002b c 1 t u002b c 2 . Essential in any modeling problem is the consistent choice of units . We measure time t in seconds and distance in meters. y ue0a2 0 ue0a3 u003d 0 gives c 2 u003d 0, and y u2032 ue0a2 0 ue0a3 u003d c 1 u003d 6. Hence y u003d 1 2 kt 2 u002b 6 t . Now the plane needs 1 min u003d 60 sec for 3 km u003d 3000 m. Hence

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2 Ordinary Differential Equations (ODEs) Part A 3000 u003d 1 2 k u2219 60 2 u002b 6 u2219 60, 1800 k u003d 3000 u2212 360, k u003d 1.47, so that v u003d y u2032 u003d 1.47 t u002b c 1 u003d 1.47 t u002b 6. Hence the answer is v ue0a2 60 ue0a3 u003d y u2032 ue0a2 60 ue0a3 u003d 1.47 u2219 60 u002b 6 u003d 94 ue0a4 m/sec ue0a5 u003d 338 ue0a4 km/h ue0a5 u003d 210 ue0a4 mph ue0a5 . Sec . 1 . 2 Geometric Meaning of y u2032 u003d f ue0a2 x , y ue0a3 . Direction Fields Problem Set 1 . 2 . Page 11 3 . Direction field , verification of solution . You may verify by differentiation that the general solution is y u003d tan ue0a2 x u002b c ue0a3 and the particular solution satisfying y ue0a2 1 4 u03c0 ue0a3 u003d 1 is y u003d tan x . Indeed, for the particular solution you obtain y u2032 u003d 1 cos 2 x u003d sin 2 x u002b cos 2 x cos 2 x u003d 1 u002b tan 2 x u003d 1 u002b y 2 and for the general solution the corresponding formula with x replaced by x u002b c .
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