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 = x 3 . The method of Section 1.5 then yields the general solution y = x3(C + ln x). 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). 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]. If we write the equation in the form y + (2 / x) y = 1/ x 3 we see that it is linear with integrating factor = x 2 . The method of Section 1.5 then yields the general solution y = x2(C + ln x). 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. 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). 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). This is a linear differential equation with integrating factor = e3 x . The method of Section 1.5 yields the general solution y = (x3 + C)e-3x. We note that Dy e x + y e x y
3.
5.
7.
9.
13.
15.
17.
)=
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. 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). Review Problems 1
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 = x + 1. The method of Section then 1.5 yields the general solution y = [C + ln(x 1)] / (x + 1). 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
23.
25.
If we write the equation in the form y + ( 2 /( x + 1) ) y = 3 we see that it is linear with integrating factor = ( x + 1) . The method of Section 1.5 then yields the general solution y = x + 1 + C (x + 1)2.
2
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 = x1(C + ln x)1/3. 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 = ( 2 x + 1) . The method of Section 1.5 then yields the general solution y = (x2 + x + C)(2x + 1)1/2.
1/ 2
29.
31. 33. 35.
dy /( y + 7) = 3x 2 dx is separable; y + 3x 2 y = 21x 2 is linear. (3x 2 + 2 y 2 ) dx + 4 xy dy = 0 is exact; y = 1 4 (3 x / y + 2 y / x ) is homogeneous. dy /( y + 1) = 2 x dx / x 2 + 1 is separable; y 2 x /( x 2 + 1) y = 2 x /( x 2 + 1) is linear.
