Part 2 Separation of Variables-Advanced Engineering Mathematics-Solution Manual, Exercises of Engineering Mathematics

Download Part 2 Separation of Variables-Advanced Engineering Mathematics-Solution Manual and more Exercises Engineering Mathematics in PDF only on Docsity! 4. Set y  9x
v. Then y
v  9x. By substitution into the given ODE you obtain y
v  9
v2. By separation,
dx. Integration gives arctan
x  c*, arctan
3x  c and from this and substitution of y
v  9x, v
3 tan (3x  c), y
3 tan (3x  c)  9x. 6. Set u
y/x. Then y
xu, y
u  xu. Substitution into the ODE and subtraction of u on both sides gives y

u  xu
 u, xu
. Separation of variables and replacement of u with y/x yields 2u du
dx, u2
8 ln x  c, y2
x2 (8 ln x  c). 8. u
y/x, y
xu, y
u  xu. Substitute u into the ODE, drop xu on both sides, and divide by x2 to get xy
xu  x2u
1_2x 2u2  xu, u
1_2u 2. Separate variables, integrate, and solve algebraically for u:
1_2 dx, 
1_ 2(x  c*), u
. Hence y
xu
. 10. By separation, y dy
4x dx. By integration, y2
4x2  c. The initial condition y(0)
3, applied to the last equation, gives 9
0  c. Hence y2  4x2
9. 12. Set u
y/x. Then y
u  xu. Divide the given ODE by x2 and substitute u and u into the resulting equation. This gives 2u(u  xu)
3u2  1. Subtract 2u2 on both sides and separate the variables. This gives 2xuu
u2  1,
. Integrate, take exponents, and then take the square root: ln (u2  1)
ln x  c*, u2  1
cx, u

cx  1. Hence y
xu
x
cx  1. From this and the initial condition, y(1)

c  1
2, c
5. This gives the answer y
x
5x  1. dx x 2u du u2  1 2x c  x 2 c  x 1 u du u2 8 x 4 u 4 u y x 4x y v 3 v 3 1 3 dv v2  9 Instructor’s Manual 5 im01.qxd 9/21/05 10:17 AM Page 5 docsity.com 14. Set u
y/x. Then y
xu, y
u  xu. Substitute this into the ODE, subtract u on both sides, simplify algebraically, and integrate: xu
cos (x2) uu
2x cos (x2), u2/2
sin (x2)  c. Hence y2
2x2(sin (x2)  c). By the initial condition, 
 (sin 1_2  c), c
0, y
xu
x
2 sin (x2). Problem Set 1.3. Problem 14. First five real branches of the solution 16. u
y/x, y
xu, y
u  xu
u  4x4 cos2u. Simplify, separate variables, and integrate: u
4x3 cos2u, du/cos2u
4x3 dx, tan u
x4  c. Hence y
xu
x arctan (x4  c). From the initial condition, y(2)
2 arctan (16  c)
0, c
16. Answer: y
x arctan (x4  16). 18. Order terms: (1  b cos )
br sin . Separate variables and integrate:
d , ln r
ln (1  b cos )  c*. b sin 1  b cos dr r dr d y x0 1 2 3 4–1–2–3–4 1 –1 –2 –3 –4 –5 –6 2 3 4 5 6 2x2 u 6 Instructor’s Manual im01.qxd 9/21/05 10:17 AM Page 6 docsity.com SECTION 1.4. Exact ODEs. Integrating Factors, page 19 Purpose. This is the second “big” method in this chapter, after separation of variables, and also applies to equations that are not separable. The criterion (5) is basic. Simpler cases are solved by inspection, more involved cases by integration, as explained in the text. Comment on Condition (5) Condition (5) is equivalent to (6) in Sec. 10.2, which is equivalent to (6) in the case of two variables x, y. Simple connectedness of D follows from our assumptions in Sec. 1.4. Hence the differential form is exact by Theorem 3, Sec. 10.2, part (b) and part (a), in that order. Method of Integrating Factors This greatly increases the usefulness of solving exact equations. It is important in itself as well as in connection with linear ODEs in the next section. Problem Set 1.4 will help the student gain skill needed in finding integrating factors. Although the method has somewhat the flavor of tricks, Theorems 1 and 2 show that at least in some cases one can proceed systematically—and one of them is precisely the case needed in the next section for linear ODEs. SOLUTIONS TO PROBLEM SET 1.4, page 25 2. (x  y) dx  (y  x) dy
0. Exact; the test gives 1 on both sides. Integrate x  y over x: u
1_2x 2  xy  k(y). Differentiate this with respect to y and compare with N: uy
x  k
y  x. Thus k
y, k
1_ 2 y 2  c*. Answer: 1_2x 2  xy  1_2 y 2
1_2(x  y) 2
c; thus y
x  c
. 4. Exact; the test gives ey  ex on both sides. Integrate M with respect to x to get u
xey  yex  k(y). Differentiate this with respect to y and equate the result to N: uy
xe y  ex  k
N
xey  ex. Hence k
0, k
const. Answer: xey  yex
c. 6. Exact; the test gives ex sin y on both sides. Integrate M with respect to x: u
ex cos y  k(y). Differentiate: uy
e x sin y  k. Equate this to N
ex sin y. Hence k
0, k
const. Answer: ex cos y
c. 8. Exact; 1/x2  1/y2 on both sides of the equation. Integrate M with respect to x: u
x2    k(y). Differentiate this with respect to y and equate the result to N: uy
   k
N, k
2y, k
y 2. Answer: x2    y2
c. y x x y 1 x x y2 y x x y Instructor’s Manual 9 im01.qxd 9/21/05 10:17 AM Page 9 docsity.com 10. Exact; the test gives 2x sin (x2) on both sides. Integrate N with respect to y to get u
y cos (x2)  l(x). Differentiate this with respect to x and equate the result to M: ux
2xy sin (x 2)  l
M
2xy sin (x2), l
0. Answer: y cos (x2)
c. 12. Not exact. Try Theorem 1. In R you have Py  Qx
e xy  1  exy(x  1)
xexy  1
Q so that R
1, F
e
x, and the exact ODE is (ey  ye
x) dx  (xey  e
x) dy
0. The test gives ey  e
x on both sides of the equation. Integration of M
FP with respect to x gives u
xey  ye
x  k(y). Differentiate this with respect to y and equate it to N
FQ: uy
xe y  e
x  k
N
xey  e
x. Hence k
0. Answer: xey  ye
x
c. 14. Not exact; 2y  y. Try Theorem 1; namely, R
(Py  Qx) /Q
(2y  y) /(xy)
3/x. Hence F
1/x 3. The exact ODE is (x  ) dx  dy
0. The test gives 2y/x3 on both sides of the equation. Obtain u by integrating N
FQ with respect to y: u
  l(x). Thus ux
 l
M
x  . Hence l
x, l
x2/2, y2/2x2  x2/2
c*. Multiply by 2 and use the initial condition y(2)
1: x2 
c
3.75 because inserting y(2)
1 into the last equation gives 4  0.25
3.75. 16. The given ODE is exact and can be written as d(cos xy)
0; hence cos xy
c, or you can solve it for y by the usual procedure. y(1)
 gives 1
c. Answer: cos xy
1. 18. Try Theorem 2. You have R*
(Qx  Py) /P
[ cos xy  x sin xy  (x sin xy  )]P
. Hence F*
y. This gives the exact ODE (y cos xy  x) dx  (y  x cos xy) dy
0. 1 y x y2 1 y y2 x2 y2 x3 y2 x3 y2 2x2 y x2 y2 x3 10 Instructor’s Manual im01.qxd 9/21/05 10:17 AM Page 10 docsity.com In the test, both sides of the equation are cos xy  xy sin xy. Integrate M with respect to x: u
sin xy  1_2x 2  k(y). Hence uy
x cos xy  k(y). Equate the last equation to N
y  x cos xy. This shows that k
y; hence k
y2/2. Answer: sin xy  1_2x 2  1_2y 2
c. 20. Not exact; try Theorem 2: R*
(Qx  Py) /P
[1  (cos 2 y  sin2 y  2x cos y sin y)] /P
[2 sin2 y  2x cos y sin y] /P
2(sin y)(sin y  x cos y) /(sin y cos y  x cos2 y)
2(sin y) /cos y
2 tan y. Integration with respect to y gives 2 ln (cos y)
ln (1/cos2 y); hence F*
1/cos2 y. The resulting exact equation is (tan y  x) dx  dy
0. The exactness test gives 1/cos2 y on both sides. Integration of M with respect to x yields u
x tan y  1_2x 2  k(y). From this, uy
 k. Equate this to N
x/cos2 y to see that k
0, k
const. Answer: x tan y  1_2x 2
c. 22. (a) Not exact. Theorem 2 applies and gives F*
1/y from R*
(Qx  Py) /P
(0  cos x) /(y cos x)
 . Integrating M in the resulting exact ODE cos x dx  dy
0 with respect to x gives u
sin x  k(y). From this, uy
k
N
. Hence k
1/y. Answer: sin x  1/y
c. Note that the integrating factor 1/y could have been found by inspection and by the fact that an ODE of the general form ƒ(x) dx  g(y) dy
0 is always exact, the test resulting in 0 on both sides. (b) Yes. Separation of variables gives dy/y2
cos x dx. By integration, 1/y
sin x  c* in agreement with the solution in (a). (d) seems better than (c). But this may depend on your CAS. In (d) the CAS may draw vertical asymptotes that disturb the figure. From the solution in (a) or (b) the student should conclude that for each nonzero y(x0)
y0 there is a unique particular solution because sin x0  1/y0
c. 1 y2 1 y2 1 y x cos2 y x cos2 y Instructor’s Manual 11 im01.qxd 9/21/05 10:17 AM Page 11 docsity.com