SECTION 5.2

GENERAL SOLUTIONS OF LINEAR EQUATIONS
Students should check each of Theorems 1 through 4 in this section to s...
y(x) = c1y1(x) + c2y2(x) + c3y3(x),

then calculate y′(x) and y″(x), and finally impose the given initial conditions to de...
c1 + c2 = 3,   3c3 = − 1,    − 9c2 = 2

       with solution c1 = 29 / 9, c2 = − 2 / 9, c3 = − 1/ 3. Hence the desired par...
22.   Imposition of the initial conditions y (0) = 0, y′(0) = 10 on the general solution
       y ( x) = c1e 2 x + c2e −2 ...
29.   If c0erx + c1xerx + ⋅⋅⋅ + cnxnerx = 0, then division by erx yields

                             c0 + c1x + ⋅⋅⋅ + cn...
Upcoming SlideShare
Loading in …5
×

Sect5 2

576 views

Published on

Published in: Technology, Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
576
On SlideShare
0
From Embeds
0
Number of Embeds
5
Actions
Shares
0
Downloads
8
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Sect5 2

  1. 1. SECTION 5.2 GENERAL SOLUTIONS OF LINEAR EQUATIONS Students should check each of Theorems 1 through 4 in this section to see that, in the case n = 2, it reduces to the corresponding theorem in Section 5.1. Similarly, the computational problems for this section largely parallel those for the previous section. By the end of Section 5.2 students should understand that, although we do not prove the existence-uniqueness theorem now, it provides the basis for everything we do with linear differential equations. The linear combinations listed in Problems 1-6 were discovered "by inspection" — that is, by trial and error. 1. (5/2)(2x) + (-8/3)(3x2) + (-1)(5x - 8x2) = 0 2. (-4)(5) + (5)(2 - 3x2) + (1)(10 + 15x2) = 0 3. (1)(0) + (0)(sin x) + (0)(ex) = 0 4. (1)(17) + (-17/2)(2 sin2x) + (-17/3)(3 cos2x) = 0, because sin2x + cos2x = 1. 5. (1)(17) + (-34)(cos2x) + (17)(cos 2x) = 0, because 2 cos2x = 1 + cos 2x. 6. (-1)(ex) + (1)(cosh x) + (1)(sinh x) = 0, because cosh x = (ex + e-x)/2 and sinh x = (ex - e-x)/2. 1 x x2 7. W = 0 1 2 x = 2 is nonzero everywhere. 0 0 2 ex e2 x e3 x 8. W = ex 2e 2 x 3e 3 x = 2 e 6 x is never zero. ex 4e 2 x 9e 3 x 9. W = ex(cos2x + sin2x) = ex ≠ 0 10. W = x-7ex(x + 1)(x + 4) is nonzero for x > 0. 11. W = x3e2x is nonzero if x ≠ 0. 12. W = x-2[2 cos2(ln x) + 2 sin2(ln x)] = 2x-2 is nonzero for x > 0. In each of Problems 13-20 we first form the general solution
  2. 2. y(x) = c1y1(x) + c2y2(x) + c3y3(x), then calculate y′(x) and y″(x), and finally impose the given initial conditions to determine the values of the coefficients c1, c2, c3. 13. Imposition of the initial conditions y (0) = 1, y ′(0) = 2, y ′′(0) = 0 on the general solution y ( x ) = c1e x + c2e − x + c3e −2 x yields the three equations c1 + c2 + c3 = 1, c1 − c2 − 2c3 = 2, c1 + c2 + 4c3 = 0 with solution c1 = 4 / 3, c2 = 0, c3 = − 1/ 3. Hence the desired particular solution is given by y(x) = (4ex - e-2x)/3. 14. Imposition of the initial conditions y (0) = 0, y ′(0) = 0, y′′(0) = 3 on the general solution y ( x ) = c1e x + c2e 2 x + c3e3 x yields the three equations c1 + c2 + c3 = 1, c1 + 2c2 + 3c3 = 2, c1 + 4c2 + 9c3 = 0 with solution c1 = 3 / 2, c2 = − 3, c3 = 3 / 2. Hence the desired particular solution is given by y(x) = (3ex - 6e2x + 3e3x)/2. 15. Imposition of the initial conditions y (0) = 2, y′(0) = 0, y′′(0) = 0 on the general solution y ( x ) = c1e x + c2 x e x + c3 x 2 e3 x yields the three equations c1 = 2, c1 + c2 = 0, c1 + 2c2 + 2c3 = 0 with solution c1 = 2, c2 = − 2, c3 = 1. Hence the desired particular solution is given by y(x) = (2 - 2x + x2)ex. 16. Imposition of the initial conditions y (0) = 1, y ′(0) = 4, y ′′(0) = 0 on the general solution y ( x ) = c1e x + c2e 2 x + c3 x e 2 x yields the three equations c1 + c2 = 1, c1 + 2c2 + c3 = 4, c1 + 4c2 + 4c3 = 0 with solution c1 = − 12, c2 = 13, c3 = − 10. Hence the desired particular solution is given by y(x) = -12ex + 13e2x - 10xe2x. 17. Imposition of the initial conditions y (0) = 3, y′(0) = − 1, y′′(0) = 2 on the general solution y ( x ) = c1 + c2 cos 3x + c3 sin 3x yields the three equations
  3. 3. c1 + c2 = 3, 3c3 = − 1, − 9c2 = 2 with solution c1 = 29 / 9, c2 = − 2 / 9, c3 = − 1/ 3. Hence the desired particular solution is given by y(x) = (29 - 2 cos 3x - 3 sin 3x)/9. 18. Imposition of the initial conditions y (0) = 1, y ′(0) = 0, y′′(0) = 0 on the general solution y ( x) = e x (c1 + c2 cos x + c3 sin x ) yields the three equations c1 + c2 = 1, c1 + c2 + c3 = 0, c1 + 2c3 = 0 with solution c1 = 2, c2 = − 1, c3 = − 1. Hence the desired particular solution is given by y(x) = ex(2 - cos x - sin x). 19. Imposition of the initial conditions y (1) = 6, y ′(1) = 14, y ′′(1) = 22 on the general solution y ( x) = c1 x + c2 x 2 + c3 x 3 yields the three equations c1 + c2 + c3 = 6, c1 + 2c2 + 3c3 = 14, 2c2 + 6c3 = 22 with solution c1 = 1, c2 = 2, c3 = 3. Hence the desired particular solution is given by y(x) = x + 2x2 + 3x3. 20. Imposition of the initial conditions y (1) = 1, y ′(1) = 5, y ′′(1) = − 11 on the general solution y ( x ) = c1 x + c2 x −2 + c3 x −2 ln x yields the three equations c1 + c2 = 1, c1 − 2c2 + c3 = 5, 6c2 − 5c3 = − 11 with solution c1 = 2, c2 = − 1, c3 = 1. Hence the desired particular solution is given by y(x) = 2x - x-2 + x-2 ln x. In each of Problems 21-24 we first form the general solution y(x) = yc(x) + yp(x) = c1y1(x) + c2y2(x) + yp(x), then calculate y′(x), and finally impose the given initial conditions to determine the values of the coefficients c1 and c2. 21. Imposition of the initial conditions y (0) = 2, y′(0) = − 2 on the general solution y ( x ) = c1 cos x + c2 sin x + 3 x yields the two equations c1 = 2, c2 + 3 = − 2 with solution c1 = 2, c2 = − 5. Hence the desired particular solution is given by y(x) = 2 cos x - 5 sin x + 3x.
  4. 4. 22. Imposition of the initial conditions y (0) = 0, y′(0) = 10 on the general solution y ( x) = c1e 2 x + c2e −2 x − 3 yields the two equations c1 + c2 − 3 = 0, 2c1 − 2c2 = 10 with solution c1 = 4, c2 = − 1. Hence the desired particular solution is given by y(x) = 4e2x - e-2x - 3. 23. Imposition of the initial conditions y (0) = 3, y′(0) = 11 on the general solution y ( x ) = c1e − x + c2 e3 x − 2 yields the two equations c1 + c2 − 2 = 3, − c1 + 3c2 = 11 with solution c1 = 1, c2 = 4. Hence the desired particular solution is given by y(x) = e-x + 4e3x - 2. 24. Imposition of the initial conditions y (0) = 4, y′(0) = 8 on the general solution y ( x ) = c1e x cos x + c2 e x cos x + x + 1 yields the two equations c1 + 1 = 4, c1 + c2 + 1 = 8 with solution c1 = 3, c2 = 4. Hence the desired particular solution is given by y(x) = ex(3 cos x + 4 sin x) + x + 1. 25. L[y] = L[y1 + y2] = L[y1] + L[y2] = f + g 26. (a) y1 = 2 and y2 = 3x (b) y = y1 + y2 = 2 + 3x 27. The equations c1 + c2x + c3x2 = 0, c2 + 2c3x + 0, 2c3 = 0 (the latter two obtained by successive differentiation of the first one) evidently imply — by substituting x = 0 — that c1 = c2 = c3 = 0. 28. If you differentiate the equation c0 + c1 x + c2 x 2 + + cn x n = 0 repeatedly, n times in succession, the result is the system c0 + c1 x + c2 x 2 + + cn x n = 0 c1 + 2c2 x + + ncn x n −1 = 0 (n − 1)!cn −1 + n !cn x = 0 n !cn = 0 of n+1 equations in the n+1 coefficients c0 , c1 , c2 , , cn . Upon substitution of x = 0, the (k+1)st of these equations reduces to k !ck = 0, so it follows that all these coefficients must vanish.
  5. 5. 29. If c0erx + c1xerx + ⋅⋅⋅ + cnxnerx = 0, then division by erx yields c0 + c1x + ⋅⋅⋅ + cnxn = 0, so the result of Problem 28 applies. 30. When the equation x2y″ - 2xy′ + 2y = 0 is rewritten in standard form y″ + (-2/x)y′ + (2/x2)y = 0, the coefficient functions p1(x) = -2/x and p2(x) = 2/x2 are not continuous at x = 0. Thus the hypotheses of Theorem 3 are not satisfied. 31. (a) Substitution of x = a in the differential equation gives y′′(a) = − p y′(a) − q(a). (b) If y(0) = 1 and y′(0) = 0, then the equation y″ - 2y′ - 5y = 0 implies that y″(0) = 2y′(0) + 5y(0) = 5. 32. Let the functions y1, y2, ⋅⋅⋅ , yn be chosen as indicated. Then evaluation at x = a of the (k - 1)st derivative of the equation c1y1 + c2y2 + ⋅⋅⋅ cnyn = 0 yields ck = 0. Thus c1 = c2 = ⋅⋅⋅ = cn = 0, so the functions are linearly independent. 33. This follows from the fact that 1 1 1 a b c = (b − a)(c − b)(c − a). a2 b2 c2 34. W(f1, f2, ⋅⋅⋅, fn) = V exp(rix), and neither V nor exp(rix) vanishes.

×