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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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1. Write the equation in the form ydy = xdx. Integrating the left side with respect to y and the right side with respect to x yields
23
y x 2 3
 =  + C, or 3y - 2x = c.
23
4. For y Ο 3/2 multiply both sides of the equation by
3 + 2y to get the separated equation (3+2y)dy = (3x 1)dx. Integration then yields
Section 2.2
9
23 3y + y = x - x + c.
6. We need x Ο 0 and | y | < 1 for this problem to be
defined. Separating the variables we get (1-y2)"1/2iy =
x 1dx. Integrating each side yields arcsiny = ln|x|+c, so y = sin[ln|x|+c], x Ο 0 (note that |y | < 1). Also, y = ± 1 satisfy the D.E., since both sides are zero.
10a. Separating the variables we get ydy = (1-2x)dx, so 2
y2
 = x - x + c. Setting x = 0 and y = -2 we have 2 = c
2
and thus y2 = 2x - 2x2 + or y = -\j 2x - 2x2 + 4 . The negative square root must be used since y(0) = -2.
10c. Rewriting y(x) as -δ/2(2-x)(x+1) , we see that y is
defined for -1 < x < 2, However, since y' does not exist for x = -1 or x = 2, the solution is valid only for the open interval -1 < x < 2.
13. Separate variables by factoring the denominator of the
2x
right side to get ydy = ------dx. Integration yields
1+x2
22
y /2 = ln(1+x )+c and use of the I.C. gives c = 2. Thus
2 1/2
y = ± [2ln(1+x ) +4] , but we must discard the plus
square root because of the I.C. Since 1 + x > 0, the solution is valid for all x.
15. Separating variables and integrating yields 22
y + y = x + c. Setting y = 0 when x = 2 yields c = -4 22
or y + y = x -4. To solve for y complete the square on
the left side by adding 1/4 to both sides. This yields
2 1 2 1 1 2 2 y + y +  = x - 4 +  or (y + ) = x - 15/4. Taking
4 4 2
the square root of both sides yields
y +  = +Vx2 - 15/4 , where the positive square root must be taken in order to satisfy the I.C. Thus
1 Γ~2 2
y = - + yx - 15/4 , which is defined for x > 15/4 or
2
x > δ/15 /2. The possibility that x < -σ/15 /2 is discarded due to the I.C.
2x
17a. Separating variables gives (2y-5)dy = (3x -e )dx and
10
Section 2.2
2 3 x
integration then gives y  5y = x  e + c. Setting x = 0 and y = 1 we have 1  5 = 0  1 + c, or c = 3.
2 3 x
Thus y  5y  (x e 3) = 0 and using the quadratic formula then gives
5 ± V 25+4(x3ex 3) 5 Γ―3 3 7
y(x) = --------------------- =------\  + x3 ex . The
2 2 V 4
negative square root is chosen due to the I.C.
17c. The interval of definition for y must be found
numerically. Approximate values can be found by plotting
13 3 x
y-, (x) = - + x and y2 (x) = e and noting the values of x
1 4 2
where the two curves cross.
11
to get sin3y = cos2x + c. Setting y = p/3 when 32
1
x = p/2 (from the I.C.) we find that 0 = - + c or
2
1 1 1 1 2
c = , so that sin3y = cos2x +  = cos x (using the
2 3 2 2
appropriate trigonometric identity). To solve for y we
must choose the branch that passes through the point
(p/2,p/3) and thus 3y = p - arcsin(3cos2x), or
p 1 2 y =  - arcsin(3cos x).
33
19c. The solution in part a is defined only for
0 < 3cos2x < 1, or  V1/3 < cosx < V1/3 . Taking the
indicated square roots and then finding the inverse cosine of each side yields .9553 < x < 2.1863, or Ixp/2 I < 0.6155, as the approximate interval.
2 2 3 2
21. We have (3y -6y)dy = (1 + 3x )dx so that y -3y =
x + x3 - 2, once the I.C. are used. From the D.E., the integral curve will have a vertical tangent when
2
3y - 6y = 0, or y = 0,2. For y = 0 we have x3 + x - 2 = 0, which is satisfied for x = 1, which is the only zero of the function w = x3 + x - 2. Likewise, for y = 2, x = -1.
2
23. Separating variables gives y dy = (2+x)dx, so
Section 2.2
11
2
-1 x
-y = 2x +  + c. y(0) = 1 yields c = -1 and thus
2
-1 2
y = -------------- = ------------. This gives
22 x2 2 - 4x - x2
 + 2x - 1
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