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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chap . 1 First- Order ODEs Sec . 1 . 1 Basic Concepts . Modeling Problem Set 1 . 1 . Page 8 1 . Calculus . To solve y u U f u x U U u sin u x is a problem of calculus, namely, to integrate u sin u x , obtaining (chain rule!) y U 1 u cos u x ¡ c . The constant of integration c is arbitrary. This is essential. It means that the ODE y u U u sin u 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 U 0, u 1,1,1/2. 13 . Initial value problem ( IVP ). Differentiation of y U ce u x 2 gives y u U u 2 cxe u x 2 . Hence the left side of the ODE becomes y u ¡ 2 xy U u 2 cxe u x 2 ¡ 2 x u ce u x 2 U U 0. This verifies that y U ce u x 2 is a solution of the given ODE. For x U 1 it has the value y u 1 U U ce u 1 . This should equal the given initial value y u 1 U U 1/ e . Hence c U 1, and the answer (the solution of the IVP) is y U e u x 2 . 17 . Modeling : Falling body . y uu U g U const is the model of the problem, an ODE of second order. Integrate to get the velocity v U y u U gt ¡ c 1 ( c 1 arbitrary). Integrate once more to obtain the distance fallen y U 1 2 gt 2 ¡ c 1 t ¡ c 2 ( c 2 arbitrary). All this is calculus. From this we obtain y U 1 2 gt 2 by imposing the initial conditions y u U U c 2 U 0 and v u U U y u u U U c 1 U 0, meaning that the stone should start from the initial position y U 0 with initial velocity 0. 19 . Airplane takeoff . The solution is given on p. A4. The acceleration is y uu U k with unknown k . Two integrations give the speed v U y u U kt ¡ c 1 and the distance y U 1 2 kt 2 ¡ c 1 t ¡ c 2 . Essential in any modeling problem is the consistent choice of units . We measure time t in seconds and distance in meters. y u U U 0 gives c 2 U 0, and y u u U U c 1 U 6. Hence y U 1 2 kt 2 ¡ 6 t . Now the plane needs 1 min U 60 sec for 3 km U 3000 m. Hence 2 Ordinary Differential Equations (ODEs) Part A 3000 u 1 2 k u 60 2 U 6 u 60, 1800 k u 3000 U 360, k u 1.47, so that v u y ¡ u 1.47 t U c 1 u 1.47 t U 6. Hence the answer is v u 60 U u y ¡ u 60 U u 1.47 u 60 U 6 u 94 ¡ m/sec ¢ u 338 ¡ km/h ¢ u 210 ¡ mph ¢ . Sec . 1 . 2 Geometric Meaning of y ¡ u f u x , y U . 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 u tan u x U c U and the particular solution satisfying y u 1 4 u U u 1 is y u tan x . Indeed, for the particular solution you obtain y ¡ u 1 cos 2 x u sin 2 x U cos 2 x cos 2 x u 1 U tan 2 x u 1 U y 2 and for the general solution the corresponding formula with x replaced by x U c . Sec . 1 . 2 . Prob . 3 . Direction Field 19 . Initial value problem . In this section the usual notation is (1), that is, y ¡ u f u x , y U , and the direction field lies in the xy-plane. In Prob. 19 the ODE is v ¡ u f u t , v U u g U bv 2 / m , where v suggests velocity....
View Full Document

This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online