1. CHAPTER 1 Review Problems
The main objective of this set of review problems is practice in the identification of the different
types of first-order differential equations discussed in this chapter. In each of Problems 1-36 we
identify the type of the given equation and indicate an appropriate method of solution.
1. If we write the equation in the form y′ − (3 / x ) y = x 2 we see that it is linear with
integrating factor ρ = x −3 . The method of Section 1.5 then yields the general solution
y = x3(C + ln x).
2. We write this equation in the separable form y′ / y 2 = ( x + 3) / x 2 . Then separation of
variables and integration as in Section 1.4 yields the general solution
y = x / (3 – Cx – x ln x).
3. This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the general solution y = x/(C – ln x).
4. (
We note that Dy 2 xy 3 + e x )= ( )
Dx 3 x 2 y 2 + sin y = 6 xy 2 , so the given equation is
exact. The method of Example 9 in Section 1.6 yields the implicit general solution
x2y3 + ex – cos y = C.
5. We write this equation in the separable form y′ / y 2 = (2 x − 3) / x 4 . Then separation
of variables and integration as in Section 1.4 yields the general solution
y = C exp[(1 – x)/x3].
6. We write this equation in the separable form y′ / y 2 = (1 − 2 x) / x 2 . Then separation
of variables and integration as in Section 1.4 yields the general solution
y = x / (1 + Cx + 2x ln x).
7. If we write the equation in the form y′ + (2 / x) y = 1/ x 3 we see that it is linear with
integrating factor ρ = x 2 . The method of Section 1.5 then yields the general solution
y = x–2(C + ln x).
8. This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the general solution y = 3Cx/(C – x3).
9. If we write the equation in the form y′ + (2 / x) y = 6 x y we see that it is a Bernoulli
equation with n = 1/2. The substitution v = y −1/ 2 of Eq. (10) in Section 1.6 then
yields the general solution y = (x2 + C/x)2.
10. ( )
We write this equation in the separable form y′ / 1 + y 2 = 1 + x 2 . Then separation
of variables and integration as in Section 1.4 yields the general solution
Chapter 1 Review 1

2. y = tan(C + x + x3/3).
11. This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the general solution y = x / (C – 3 ln x).
12. ( ) ( )
We note that Dy 6 xy 3 + 2 y 4 = Dx 9 x 2 y 2 + 8 xy 3 = 18 xy 2 + 8 y 3 , so the given
equation is exact. The method of Example 9 in Section 1.6 yields the implicit general
solution 3x2y3 + 2xy4 = C.
13. We write this equation in the separable form y′ / y 2 = 5 x 4 − 4 x. Then separation
of variables and integration as in Section 1.4 yields the general solution
y = 1 / (C + 2x2 – x5).
14. This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the implicit general solution y2 = x2 / (C + 2 ln x).
15. This is a linear differential equation with integrating factor ρ = e3 x . The method of
Section 1.5 yields the general solution y = (x3 + C)e-3x.
16. The substitution v = y − x, y = v + x, y′ = v′ + 1 gives the separable equation
v′ + 1 = ( y − x) 2 = v 2 in the new dependent variable v. The resulting implicit general
solution of the original equation is y – x – 1 = C e2x(y – x + 1).
17. (
We note that Dy e x + y e x y )= (
Dx e y + x e x y )= e x y + xy e x y , so the given equation is
exact. The method of Example 9 in Section 1.6 yields the implicit general solution
ex + ey + ex y = C.
18. This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the implicit general solution y2 = Cx2(x2 – y2).
19. ( )
We write this equation in the separable form y′ / y 2 = 2 − 3x 5 / x 3 . Then separation
of variables and integration as in Section 1.4 yields the general solution
y = x2 / (x5 + Cx2 + 1).
20. If we write the equation in the form y′ + (3 / x) y = 3 x −5 / 2 we see that it is linear with
integrating factor ρ = x 3 . The method of Section 1.5 then yields the general solution
y = 2x–3/2 + Cx–3.
21. If we write the equation in the form y′ + (1/( x + 1) ) y = 1/( x 2 − 1) we see that it is linear
with integrating factor ρ = x + 1. The method of Section then 1.5 yields the general
solution y = [C + ln(x – 1)] / (x + 1).
Chapter 1 Review 2

3. 22. If we write the equation in the form y′ − (6 / x) y = 12 x 3 y 2 / 3 we see that it is a Bernoulli
equation with n = 1/3. The substitution v = y −2 / 3 of Eq. (10) in Section 1.6 then
yields the general solution y = (2x4 + Cx2)3.
23. ( ) ( )
We note that Dy e y + y cos x = Dx x e y + sin x = e y + cos x, so the given equation
is exact. The method of Example 9 in Section 1.6 yields the implicit general solution
x ey + y sin x = C
24. ( )
We write this equation in the separable form y′ / y 2 = 1 − 9 x 2 / x 3/ 2 . Then separation
of variables and integration as in Section 1.4 yields the general solution
y = x1/2 / (6x2 + Cx1/2 + 2).
25. If we write the equation in the form y′ + ( 2 /( x + 1) ) y = 3 we see that it is linear with
integrating factor ρ = ( x + 1) . The method of Section 1.5 then yields the general
2
solution y = x + 1 + C (x + 1)–2.
26. ( ) ( )
We note that Dy 9 x1/ 2 y 4 / 3 − 12 x1/ 5 y 3/ 2 = Dx 8 x 3/ 2 y1/ 3 − 15 x 6 / 5 y1/ 2 =
12 x1/ 2 y1/ 3 − 18 x1/ 5 y1/ 2 , so the given equation is exact. The method of Example 9 in
Section 1.6 yields the implicit general solution 6x3/2y4/3 – 10x6/5y3/2 = C.
27. If we write the equation in the form y′ + (1/ x) y = − x 2 y 4 / 3 we see that it is a Bernoulli
equation with n = 4. The substitution v = y −3 of Eq. (10) in Section 1.6 then yields
the general solution y = x–1(C + ln x)–1/3.
28. If we write the equation in the form y′ + (1/ x) y = 2 e 2 x / x we see that it is linear with
integrating factor ρ = x. The method of Section 1.5 then yields the general solution
y = x–1(C + e2x).
29. If we write the equation in the form y′ + (1/(2 x + 1) ) y = (2 x + 1)1/ 2 we see that it is
linear with integrating factor ρ = ( 2 x + 1) . The method of Section 1.5 then yields
1/ 2
the general solution y = (x2 + x + C)(2x + 1)–1/2.
30. The substitution v = x + y, y = v − x, y′ = v′ − 1 gives the separable equation
v′ − 1 = v in the new dependent variable v. The resulting implicit general solution of
the original equation is x = 2(x + y)1/2 – 2 ln[1 + (x + y)1/2] + C.
31. dy /( y + 7) = 3x 2 is separable; y′ + 3x 2 y = 21x 2 is linear.
32. dy /( y 2 − 1) = x is separable; y′ + x y = x y 3 is a Bernoulli equation with n = 3.
Chapter 1 Review 3

4. 33. (3x 2 + 2 y 2 ) dx + 4 xy dy = 0 is exact; y′ = − 1 (3x / y + 2 y / x ) is homogeneous.
4
1+ 3y / x
34. ( x + 3 y ) dx + (3x − y ) dy = 0 is exact; y′ = is homogeneous.
y / x−3
35. ( ) ( )
dy /( y + 1) = 2 x / x 2 + 1 is separable; y′ − 2 x /( x 2 + 1) y = 2 x /( x 2 + 1) is linear.
36. dy / ( )
y − y = cot x is separable; y′ + (cot x ) y = (cot x) y is a Bernoulli equation
with n = 1/2.
Chapter 1 Review 4