This document provides 36 review problems to identify the type of first-order differential equations and appropriate solution methods discussed in Chapter 1. The problems cover linear equations, separable equations, homogeneous equations, exact equations, and Bernoulli equations. An example solution is provided for several problems identifying the type of equation and corresponding general solution.

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