![]() ![]() ![]() , which is defined for all x provided 2x2 is never equal to 2x20 − 1/y0 this If y0 �= 0 then, proceeding as before, we get C = 2x2 − 1y, C = 2x2 The methods ofExample 4 fail because the integrals there become divergent when the point x = 0 is included inthe integral.Ĥ0. Janu11:46 L24-ch09 Sheet number 4 Page number 398 blackĢx2 − C, all pass through the point(0,− 1Ī solution of the initial value problem with y(0) = 0 is (by inspection) y = 0. Janu11:46 L24-ch09 Sheet number 3 Page number 397 black Y= −x dx, ln |y| = −x2/2 + C1, y = ±eC1e−xĢ/2 = Ce−x2/2, including C = 0 by inspectionĭx = secx tanx dx, ey = secx+ C, y = ln(secx+ C)īy inspection, y = 0 is also a solution, as is y = 1.Ģ4.1ydy = cosx dx, ln |y| = sinx+ C, y = C1e )dy = exdx, 2 ln |y|+ y2 = ex + C by inspection, y = 0 is also a solution Janu11:46 L24-ch09 Sheet number 2 Page number 396 black ![]() (b) second order y′ = c1et − c2e−t, y′′ − y = c1et + c2e−t −(c1e (b) second order y′ = c1 cos t− c2 sin t, y′′ + y = −c1 sin t− c2 cos t+ (c1 sin t+ c2 cos t) = 0 y′ = x3 − 2 sinx, y(0) = 3 by inspection. Mathematical Modeling with DifferentialEquationsġ. Janu11:46 L24-ch09 Sheet number 1 Page number 395 black µ = e 2 x dx = e x 2, d dx ye x 2 = xe x 2 ,ye x 2 = 1 2 e x 2 + C, y = 1 2 + Ce − x 2 µ = e 4 dx = e 4 x, e 4 x y = e x dx = e x + C, y = e − 3 x + Ce − 4 x 10. (a) IF: µ = e − 4 x dx = e − 2 x 2, d dx ye − 2 x 2 =0 ,y = Ce 2 x 2 separation of variables: dy y =4 x dx, ln | y | =2 x 2 + C 1 ,y = ± e C 1 e 2 x 2 = Ce 2 x 2 including C = 0 by inspection (b) IF: µ = e dt = e t, d dt ye t =0 ,y = Ce − t separation of variables: dy y = − dt, ln | y | = − t + C 1 ,y = ± e C 1 e − t = Ce − t including C = 0 by inspection 9. (a) IF: µ = e 3 dx = e 3 x, d dx ye 3 x =0 ,ye 3 x = C, y = Ce − 3 x separation of variables: dy y = − 3 dx, ln | y | = − 3 x + C 1 ,y = ± e − 3 x e C 1 = Ce − 3 x including C = 0 by inspection (b) IF: µ = e − 2 dt = e − 2 t, d dt =0 ,ye − 2 t = C, y = Ce 2 t separation of variables: dy y =2 dt, ln | y | =2 t + C 1 ,y = ± e C 1 e 2 t = Ce 2 t including C = 0 by inspection 8. ![]() 2 x + y 2 +2 xy dy dx = 0, by inspection. 1 y dy dx = y 2 +2 xy dy dx, dy dx (1 − 2 xy 2 )= y 3, dy dx = y 3 1 − 2 xy 2 6. (a) first order 2 dy dx + y =2 − c 2 e − x/ 2 +1 + ce − x/ 2 + x − 3= x − 1 (b) second order y = c 1 e t − c 2 e − t ,y − y = c 1 e t + c 2 e − t − ( c 1 e t + c 2 e − t ) =0 5. (a) first order dy dx = c (1 + x ) dy dx = (1 + x ) c = y (b) second order y = c 1 cos t − c 2 sin t, y + y = − c 1 sin t − c 2 cos t +( c 1 sin t + c 2 cos t )=0 4. y = x 3 − 2 sin x, y (0) = 3 by inspection. y =9 x 2 e x 3 =3 x 2 y and y (0) = 3 by inspection. Janu11:46 L24-ch09 Sheet number 1 Page number 395 black 395 CHAPTER 9 Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. ![]()
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