dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

Transcription

1 Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources, bu please wrie up your own soluions. Copying anoher suden s soluion and handing i in as your own is considered cheaing. Don do i! (1) Secion 2.1. Problem 10 Use MAPLE (or some oher compuer sofware) o obain a direcion field for he given differenial equaion. Prin i ou and by hand skech on he vecor field approximae soluion curves passing hrough each of he given poins: (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See nex page dx = xey 1

2 2

3 3 (2) Secion 2.1 problem 30. Consider he auonomous differenial equaion /dx = f(y) where he graph of f is given in he book (no reproduced here). Use he graph o locae he criical poins of he differenial equaion. Skech a phase porrai. And by hand, skech ypical soluion curves in he subregions in he xy-plane deermined by he graphs of he equilibrium soluions. SOLUTION: The criical values of he differenial equaion are a approximaely y = 2.2, 0.5 and y = 1.5. For he phase porrai and ypical soluion curves, see he nex page.

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5 5 (3) Secion 2.2. Problem 2: Find an explici soluion o he given iniial value problem: dx = y2 1 x 2, y(2) = 2. 1 Soluion: The soluion is y = x on he inerval (1, ). How o find his: This is a separable differenial equaion, so we separae he variables and inegrae boh sides: (1) y 2 1 = Using he parial fracions decomposiion 1 u 2 1 = 1 (u 1)(u + 1) = 1 2 dx x 2 1 ( 1 u 1 1 u + 1 We inegrae boh sides of equaion (1) o obain ( ) 1 y 1 (2) 2 ln = 1 ( ) x 1 y ln + C x + 1 I is probably easier o find C using our consrain firs, and hen solve for y in erms of x: Using y = 2 when x = 2 (our iniial value consrain) we obain 1/2(ln(1/3)) = 1/2(ln(1/3)) + C So, C = 0. So, solving (2) where C = 0 we ge (y 1)/(y + 1) = (x 1)/(x + 1). Eiher you can observe ha y = x is a soluion, or do some algebra o come o his conclusion. Now, reurning o our original expression dx = y2 1 x 2 1 we see ha he righ-hand side is coninuous and has parial derivaives coninuous for all values of (x, y) excep when x = 1 or x = 1. So he larges inerval conaining our iniial value x = 2 on which he differenial equaion is defined is (1, ). () Secion 2.3. Problem 10: Find he general soluion o he differenial equaion. Give he larges inerval I over which he general soluion is defined. x + 2y = 3. dx Soluion: y = 2x+ C x is a one parameer family of soluions. Eiher inerval 2 (, 0) or (0, ) is an appropriae inerval for he soluion. How do we find his soluion? Well, his is a linear firs-order equaion, so we can solve i by rewriing i in he sandard form dx + 2 x y = 3 x and muliplying by he appropriae inegraing facor. In his case ha is e 2 x dx = x 2. Doing his we obain he differenial equaion x 2 dx +2xy = 6x 2. Inegraing boh sides wih respec o x we obain x 2 y = 2x 3 + C. Solving for y his is y = 2x + C x. This is defined on he inervals (, 0) 2 and (0, ). Eiher inerval is a larges inerval on which he soluion is valid. )

6 6 (5) Secion 2.3, Problem 28. Solve he given iniial value problem. Give he larges inerval I over which he soluion is defined. y dx x = 2y2, y(1) = 5 Soluion: y = 9/5+ (9/5) 2 +8x on he inerval ( , ). How o ge his? Well, his equaion is no linear in y bu i is se up as a linear differenial equaion wih independen variable y and dependen variable x. So we can sill solve by firs rewriing i in he sandard form dx x/y = 2y and muliplying by he inegraing facor = e 1 y = y 1 we obain 1 dx y x y 2 = 2 Inegraing boh sides (wih respec o y, reaing x as a funcion of y) gives x y = 2y + C. Subsiuing in he iniial value o solve for C gives 1/5 = 10 + C. So C = 9/5. So we have x y = 2y 9/5, x = 2y 2 (9/5)y 2y 2 (9/5)y x = 0 So y = 9/5+ (9/5) 2 +8x (since y(1) = 5 we need o ake he posiive square roo, oherwise if y = 9/5 (9/5) 2 +8x we would have y(1) = 8/5). And he larges inerval on which his is defined is when (9/5) 2 + 8x > 0. Solving his for x gives x > (6) Secion 2.. Problem 2: Solve he given iniial-value problem. ( 3y 2 2 ) y 5 d + = 0, subjec o y(1) = 1. 2y Soluion: Once we wrie his in differenial form we can check o see if i is exac. ( 3y 2 2 ) y 5 + 2y d = 0 which is of course he same as 2y d + 3y2 2 y 5 = 0 2y, and N(, y) = 3y2 2 y 5, and i is easily checked ha and so M(, y) = N = M y = 2 y. So his is an exac equaion. 5 To solve i we inegrae M(, y) wih respec o o ge f(, y) = 2 y + g(y). 2y d =

7 7 To find g(y) we know ha f y = N(, y). Bu f y = 2 y 5 + g (y) = 2 +y 5 g (y) y. So y 5 g (y) y 5 = N(, y) = 3y2 2 y 5 from which i is clear ha g (y) = 3y 3. So g(y) = 3 2y 2. Therefore our soluion is of he form f(, y) = C, so we have 2 y 3 2y 2 = C Our iniial value consrain y(1) = 1 gives us 1/ 3/2 = C. So C = 5/ So an implici soluion o he IVP is 2 y 3 2y 2 = 5/