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Differential Equations Theory, Technique, and Practice 3rd Edition by Steven Krantz -All Chapters 1-13 | SOLUTIONS MANUAL $31.35
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Differential Equations Theory, Technique, and Practice 3rd Edition by Steven Krantz -All Chapters 1-13 | SOLUTIONS MANUAL

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SOLUTIONS MANUAL for Differential Equations Theory, Technique, and Practice 3rd Edition By Steven G. Krantz ISBN 2702. All Chapters 1-13. Table of Contents 1. What Is a Differential Eq uation? 1.1 Introductory Remarks 1.2 A Taste of Ordinary Differential Equations 1.3 The Nature of Solutions 2. So...

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  • February 2, 2023
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  • technique
  • differential equations theory
  • and practice 3rd edition by steven g
  • solutions manual for differential equations theory
  • and practice 3rd edition by steven g krantz isbn 97810321

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,
,8 CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?

(k) If y 2 = x2 −cx, then 2yy 0 = 2x−c so 2xyy 0 = 2x2 −cx = x2 +x2 −cx =
x2 + y 2 .
(l) If y = c2 + c/x, then y 0 = −c/x2 so x4 (y 0 )2 = c2 = y − c/x. Use the
fact that −c/x = xy 0 to obtain x4 (y 0 )2 = y + xy 0 .
xy 0 −y y/x xyy 0 −y 2
(m) If y = cey/x , then y 0 = x2 ce = x2 . Solve for y 0 to obtain
y 0 = y 2 /(xy − y 2 ).
(n) If y + sin y = x, then y 0 + y 0 cos y = 1 or y 0 = 1/(1 + cos y). Multiply
the numerator and denominator of the right side by y to obtain y 0 =
y/(y + y cos y). Now use the identity y = x − sin y to obtain y 0 =
(x − sin y + y cos y).
(o) If x + y = arctan y, then 1 + y 0 = y 0 /(1 + y 2 ). Consequently, (1 +
y 0 )(1 + y 2 ) = y 0 . This simplifies to 1 + y 2 + y 2 y 0 = 0.

2. Find the general solution of each of the following differential equations.

(a) If y 0 = e3x − x, then y = xex dx + C = xex − ex + C.
R

2 2 x2
(b) If y 0 = xex , then y = xex dx + C = e2 + C (substitution u = x2 ).
R

(c) If (1 + x)y 0 = x, then y 0 = 1+x x
R x R
and y = 1+x dx + C = (1 −
1
1+x )dx + C = x − ln |1 + x| + C.

(d) If (1 + x2 )y 0 = x, then y 0 = 1+x x
R x 1
2 and y = 1+x2 dx + C = 2 ln |1 +
2 2
x | + C (substitution u = 1 + x ).
(e) If (1 + x2 )y 0 = arctan x, then y 0 = arctan x
R arctan x
1+x2 and y = 1+x2 dx + C =
1 x
2 arctan +C (substitution u = arctan x).
(f) y 0 = 1/x, so y = x1 dx + C = ln |x| + C.
R
R √
(g) y = arcsin xdx + C = x arcsin x + 1 − x2 + C (integrate by parts,
u = arcsin x).
(h) y 0 = sin1 x so y = sin1 x dx+C = csc xdx+C = csc x csc csc x+cot x
R R R
x+cot x dx+
C. SetR u = csc x + cot x, then du = (− csc x cot x − csc2 x)dx and
y = − du u + C = − ln |u| + C = − ln | csc x + cot x| + C.
1 1
3 x+ 3
R −1/3
(i) y 0 = 1+x x
R x
3 , so y = 1+x3 dx = x+1 + x2 −x+1 dx (partial frac-
R 1 x− 16 R 1/2
tions) = − 31 ln |x + 1| + x23 −x+1 dx + x2 −x+1 dx = − 13 ln |x + 1| +
1 1/2
2
dx = − 31 ln |x + 1| + 16 ln |x2 − x + 1| +
R
6 ln |x − x + 1| + (x− 12 )2 + 34
√
2/3
dx = − 31 ln |x + 1| + 16 ln |x2 − x + 1| + 33 arctan 2x−1
R
1+( 2x−1
√ )2
√
3
+C
3
2x−1
(substitution u = √ ).
3
−1
(j) y 0 = x2 −3x+2
x x 2
R R
, so y = x2 −3x+2 dx = x−1 + x−2 dx = − ln |x − 1| +
2 ln |x − 2| + C.

3. For each of the following differential equations, find the particular solution
that satisfies the given initial condition.

, 1.3. THE NATURE OF SOLUTIONS 9

(a) If y 0 = xex , then y = xex dx + C = (x − 1)ex + C (integrate by
R

parts, u = x). When x = 1, y = C so the particular solution is
y(x) = (x − 1)ex + 3.
(b) If y 0 = 2 sin x cos x, then y = 2 sin x cos xdx+C = sin2 x+C. When
R

x = 0, y = C so the particular solution is y(x) = sin2 x + 1.
(c) If y 0 = ln x, then y = ln xdx + C = x ln x − x + C (integrate by
R

parts, u = ln x). When x = e, y = C so the particular solution is
y(x) = x ln x − x.
(d) If y 0 = 1/(x2 − 1), then y = 1/(x2 − 1)dx + C = 1/2 1/(x −
R R

1) − 1/(x + 1)dx + C = 21 ln x−1 x+1 + C (method of partial fractions).
When x = 2, y = 12 ln 13 + C = C − ln23 so the particular solution is
y(x) = 12 ln x−1 ln 3
x+1 + 2 .

(e) If y 0 = x(x21−4) , then y = 1
R R
x(x2 −4) dx + C = 1/8 1/(x + 2) +
2
1/(x − 2) − 2/xdx + C = 18 ln |x x−4|
2 + C (method of partial fractions).
When x = 1, y = 18 ln 3 + C so the particular solution is y(x) =
2 2
1
8 ln |x x−4|
2 − 1
8 ln 3 = 1
8 ln |x3x−4|
2 .

2x2 +x 2x2 +x
(f) If y 0 = (x+1)(x 1
R R 1
2 +1) , then y = (x+1)(x2 +1) dx + C = 2 x+1 +
3x−1 1 3 2 1
x2 +1 dx + C = 2 ln(x + 1) + 4 ln(x + 1) − 2 arctan x + C (method of
partial fractions). When x = 0, y = C so the particular solution is
y(x) = 12 ln(x + 1) + 34 ln(x2 + 1) − 21 arctan x + 1.
2 Rx 2 Rx
4. By the Product Rule, the derivative of y = ex 0
e−t2 dt is 2xex 0
e−t2 dt+
2 2
ex e−x = 2xy + 1.

5. For the differential equation

y 00 − 5y 0 + 4y = 0,

carry out the detailed calculations required to verify these assertions.

(a) If y = ex , then y 00 − 5y 0 + 4y = ex − 5ex + 4ex ≡ 0.
If y = e4x , then y 00 − 5y 0 + 4y = 16e4x − 20e4x + 4e4x ≡ 0.
(b) If y = c1 ex +c2 e4x , then y 00 −5y 0 +4y = c1 (ex −5ex +4ex )+c2 (16e4x −
20e4x + 4e4x ≡ 0.

6. Taking a derivative with respect to x on both sides of x2 y = ln y + c,
0
we have 2xy + x2 y 0 = yy . Multiply y to get 2xy 2 + x2 yy 0 = y 0 . Thus
2xy 2
y0 = 1−x2 y .

7. For which values of m will the function y = ym = emx be a solution of the
differential equation

2y 000 + y 00 − 5y 0 + 2y = 0?

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