Sect1 5

SECTION 1.5

LINEAR FIRST-ORDER EQUATIONS

1. ( )
ρ = exp ∫ 1 dx = e x ; (
Dx y ⋅ e x = 2e x ; ) y ⋅ e x = 2e x + C ; y ( x) = 2 + Ce − x

y (0) = 0 implies C = − 2 so y ( x) = 2 − 2e − x

2. ρ = exp ( ∫ (−2) dx ) = e −2 x
; (
Dx y ⋅ e −2 x = 3; ) y ⋅ e −2 x = 3 x + C ; y ( x) = (3 x + C )e2 x

y (0) = 0 implies C = 0 so y ( x) = 3x e 2 x

3. ρ = exp ( ∫ 3 dx ) = e 3x
; (
Dx y ⋅ e3 x = 2 x; ) y ⋅ e3 x = x 2 + C ; y ( x ) = ( x 2 + C ) e −3 x

4. ρ = exp ( ∫ (−2 x) dx ) = e − x2
; (
Dx y ⋅ e − x
2

) = 1; y ⋅ e− x = x + C;
2
y ( x ) = ( x + C )e x
2

5. ρ = exp ( ∫ (2 / x) dx ) = e 2ln x
= x2 ; ( )
Dx y ⋅ x 2 = 3x 2 ; y ⋅ x 2 = x3 + C

y( x) = x + C / x2 ; y (1) = 5 implies C = 4 so y ( x) = x + 4 / x 2

6. ρ = exp ( ∫ (5 / x) dx ) = e 5ln x
= x5 ; ( )
Dx y ⋅ x 5 = 7 x 6 ; y ⋅ x5 = x 7 + C

y( x) = x 2 + C / x5 ; y (2) = 5 implies C = 32 so y ( x) = x 2 + 32 / x 5

7. ρ = exp ( ∫ (1/ 2 x) dx ) = e (ln x ) / 2
= x; (
Dx y ⋅ x = 5; ) y ⋅ x = 5x + C

y( x) = 5 x + C / x

8. ρ = exp ( ∫ (1/ 3x) dx ) = e (ln x ) / 3
= 3 x; (
Dx y ⋅ 3 x = 4 3 x ; ) y ⋅ 3 x = 3x4 / 3 + C

y ( x ) = 3 x + Cx −1/ 3

9. ρ = exp ( ∫ (−1/ x) dx ) = e − ln x
= 1/ x; Dx ( y ⋅1/ x ) = 1/ x; y ⋅1/ x = ln x + C

y ( x ) = x ln x + C x; y (1) = 7 implies C = 7 so y ( x) = x ln x + 7 x

10. ρ = exp ( ∫ (−3 / 2 x) dx ) = e ( −3ln x ) / 2
= x −3/ 2 ; ( )
Dx y ⋅ x −3/ 2 = 9 x1/ 2 / 2; y ⋅ x −3/ 2 = 3x 3/ 2 + C

Section 1.5 1

y ( x) = 3 x 3 + Cx 3/ 2

11. ρ = exp ( ∫ (1/ x − 3) dx ) = e ln x −3 x
= x e −3 x ; (
Dx y ⋅ x e −3 x = 0; ) y ⋅ x e −3 x = C

y ( x ) = C x −1e3 x ; y (1) = 0 implies C = 0 so y ( x) ≡ 0 (constant)

12. ρ = exp ( ∫ (3 / x) dx ) = e 3 ln x
= x3 ; (
Dx y ⋅ x3 = 2 x 7 ; ) y ⋅ x 3 = 1 x8 + C
4

y( x) = 1
4
x 5 + C x −3 ; y (2) = 1 implies C = 56 so y ( x) = 1
4
x 5 + 56 x −3

13. ( )
ρ = exp ∫ 1 dx = e x ; (
Dx y ⋅ e x = e 2 x ; ) y ⋅ e x = 1 e2 x + C
2

y ( x) = 1
2
e x + C e− x ; y (0) = 1 implies C = 1
2 so y ( x) = 1
2
e x + 1 e− x
2

14. ρ = exp ( ∫ (−3/ x) dx ) = e −3ln x
= x −3 ; (
Dx y ⋅ x −3 = x −1 ; ) y ⋅ x −3 = ln x + C

y ( x) = x 3 ln x + C x 3 ; y (1) = 10 implies C = 10 so y ( x ) = x 3 ln x + 10 x 3

15. ρ = exp ( ∫ 2 x dx ) = e x2
; (
Dx y ⋅ e x
2

) = xe x2
; y ⋅ ex = 1 ex + C
2

2
2

y ( x) = + C e− x ; y (0) = −2 implies C = − 5 so y ( x ) = − 5 e− x
2 2
1 1
2 2 2 2

16. ρ = exp ( ∫ cos x dx ) = e sin x
; ( )
Dx y ⋅ esin x = esin x cos x; y ⋅ esin x = esin x + C

y ( x ) = 1 + C e − sin x ; y (π ) = 2 implies C = 1 so y ( x ) = 1 + e− sin x

17. ( )
ρ = exp ∫ 1/(1 + x) dx = eln(1+ x ) = 1 + x; Dx ( y ⋅ (1 + x )) = cos x; y ⋅ (1 + x ) = sin x + C

C + sin x 1 + sin x
y( x) = ; y (0) = 1 implies C = 1 so y ( x ) =
1+ x 1+ x

18. ρ = exp ( ∫ (−2 / x) dx ) = e −2ln x
= x −2 ; (
Dx y ⋅ x −2 = cos x; ) y ⋅ x −2 = sin x + C

y ( x ) = x 2 (sin x + C )

19. ρ = exp ( ∫ cot x dx ) = e ln(sin x )
= sin x; Dx ( y ⋅ sin x ) = sin x cos x

y ⋅ sin x = 1 sin 2 x + C ;
2
y ( x) = 1
2 sin x + C csc x

Section 1.5 2

20. ρ = exp ( ∫ (−1 − x) dx ) = e − x − x2 / 2
; (
Dx y ⋅ e − x − x
2
/2
) = (1 + x) e − x − x2 / 2

y ⋅ e− x− x = − e− x− x + C; y ( x) = − 1 + C e − x− x
2 2 2
/2 /2 /2

y (0) = 0 implies C = 1 so y ( x ) = − 1 + e − x − x
2
/2

21. ρ = exp ( ∫ (−3 / x) dx ) = e −3ln x
= x −3 ; (
Dx y ⋅ x −3 = cos x; ) y ⋅ x −3 = sin x + C

y ( x) = x 3 sin x + C x 3 ; y (2π ) = 0 implies C = 0 so y ( x) = x 3 sin x

22. ρ = exp ( ∫ (−2 x) dx ) = e − x2
; (
Dx y ⋅ e − x
2

) = 3x ; 2
y ⋅ e− x = x3 + C
2

(
y ( x) = x 3 + C e − x ; ) y (0) = 5 implies C = 5 so y ( x ) = x 3 + 5 e − x ( )
2 2

23. ρ = exp ( ∫ (2 − 3 / x) dx ) = e 2 x −3ln x
= x −3e 2 x ; (
Dx y ⋅ x −3e 2 x = 4 e 2 x )
y ⋅ x −3 e 2 x = 2 e 2 x + C ; y ( x ) = 2 x 3 + C x 3 e −2 x

24. ρ = exp ( ∫ 3x /( x 2
)
+ 4) dx = e3ln( x
2
+ 4) / 2
= ( x 2 + 4)3/ 2 ; ( )
Dx y ⋅ ( x 2 + 4)3/ 2 = x ( x 2 + 4)1/ 2

y ⋅ ( x 2 + 4)3/ 2 = 1 ( x 2 + 4)3/ 2 + C ;
3 y ( x) = 1
3 + C ( x 2 + 4) −3/ 2

y (0) = 1 implies C = 16 so y ( x) =
3
1
3 + 16 ( x 2 + 4) −3/ 2
3

25. First we calculate

+ 3x dx = + 3x − 3x # dx =
, x + 1 , ! x + 1$
3
3 2
- 2
- 2
2
x − ln( x 2 + 1) .

It follows that ρ = ( x 2 + 1) −3/ 2 exp(3x 2 / 2) and thence that

(
Dx y ⋅ ( x 2 + 1) −3/ 2 exp(3x 2 / 2) = 6 x ( x 2 + 4)−5 / 2 , )
−3/ 2
y ⋅ ( x + 1)2
exp(3x / 2) = − 2( x 2 + 4) −3/ 2 + C ,
2

y ( x) = − 2 exp(3 x 2 / 2) + C ( x 2 + 1)3/ 2 exp(−3x 2 / 2).

Finally, y(0) = 1 implies that C = 3 so the desired particular solution is

y ( x ) = − 2 exp(3x 2 / 2) + 3( x 2 + 1)3/ 2 exp(−3 x 2 / 2).

Section 1.5 3

26. With x′ = dx / dy , the differential equation is y 3 x′ + 4 y 2 x = 1. Then with y as the
independent variable we calculate

ρ ( y ) = exp ( ∫ (4 / y) dy ) = e 4ln y
= y4; ( )
Dy x ⋅ y 4 = y

1 2 1 C
x ⋅ y4 = y + C; x( y ) = 2
+ 4
2 2y y

27. With x′ = dx / dy , the differential equation is x′ − x = y e y . Then with y as the

ρ ( y ) = exp ( ∫ (−1) dy ) = e −y
; (
Dy x ⋅ e − y )= y

x ⋅ e− y = 1
2 y 2 + C; x( y ) = ( 1
2 y2 + C ey )
28. With x′ = dx / dy , the differential equation is (1 + y 2 ) x′ − 2 y x = 1. Then with y as the

ρ ( y ) = exp ( ∫ (−2 y /(1 + y ) dy ) = e2 − ln( y 2 +1)
= (1 + y 2 )−1

( )
Dy x ⋅ (1 + y 2 )−1 = (1 + y 2 ) −2

An integral table (or trigonometric substitution) now yields

x ⌠ dy 1 y 
=  =  + tan −1 y + C 
1+ y2 ( ) 2  1+ y
2 2
⌡ 1+ y
2


x( y ) = 1
2
 y + 1 + y2
 ( )( tan −1
y+C 
 )

29. ρ = exp ( ∫ (−2 x) dx ) = e ; − x2
(
Dx y ⋅ e − x
2

)=e − x2
; y ⋅ e − x = C + ∫ e −t dt
2 x

0
2

y ( x ) = e (C + erf ( x) )
− x2 π
2

30. After division of the given equation by 2x, multiplication by the integrating factor
ρ = x–1/2 yields

x −1/ 2 y ′ − 1 x −3/ 2 y = x −1/ 2 cos x,
2

2 7
Dx x −1/ 2 y = x −1/ 2 cos x,

x −1 / 2 y = C + I
1
x
t −1/ 2 cos t dt.

Section 1.5 4

The initial condition y(1) = 0 implies that C = 0, so the desired particular solution is

y( x ) = x 1 / 2 I
1
x
t −1/ 2 cos t dt .

yc = C e ∫ (− P ) = − P yc , so yc + P yc = 0.
− P dx
31. (a) ′ ′

− P dx   − P dx
y′p = ( − P) e ∫ ⋅ ⌠  Q e ∫  dx  + e ∫ ⋅ Q e ∫
P dx P dx
(b)   = − Py p + Q
⌡   

32. (a) If y = A cos x + B sin x then

y′ + y = ( A + B) cos x + ( B − A)sin x = 2sin x

provided that A = –1 and B = 1. These coefficient values give the particular solution
yp(x) = sin x – cos x.

(b) The general solution of the equation y′ + y = 0 is y(x) = Ce–x so addition to the
particular solution found in part (a) gives y(x) = Ce–x + sin x – cos x.

(c) The initial condition y(0) = 1 implies that C = 2, so the desired particular
solution is y(x) = 2e–x + sin x – cos x.

33. The amount x(t ) of salt (in kg) after t seconds satisfies the differential equation
x′ = − x / 200, so x(t ) = 100 e− t / 200 . Hence we need only solve the equation
10 = 100 e − t / 200 for t = 461 sec = 7 min 41 sec (approximately).

34. Let x(t ) denote the amount of pollutants in the lake after t days, measured in millions of
cubic feet. Then x(t ) satisfies the linear differential equation dx / dt = 1/ 4 − x /16 with
solution x(t ) = 4 + 16 e − t /16 satisfying x(0) = 20. The value of t such that x = 8 is
t = 16 ln 4 ≈ 22.2 days. For a complete solution see Example 4 in Section 7.6 of Edwards
and Penney, Calculus with Analytic Geometry (5th edition, Prentice-Hall, 1998).

35. The only difference from the Example 4 solution in the textbook is that V = 1640 km3
and r = 410 km3/yr for Lake Ontario, so the time required is

V
t = ln 4 = 4 ln 4 ≈ 5.5452 years.
r

36. (a) The volume of brine in the tank after t min is V(t) = 60 – t gal, so the initial
value problem is

Section 1.5 5

dx 3x
= 2− , x (0 ) = 0.
dt 60 − t
The solution is
(60 − t )3
x (t ) = (60 − t ) − .
3600

(b) The maximum amount ever in the tank is 40 / 3 ≈ 23.09 lb. This occurs after
t = 60 − 20 3 ≈ 25 / 36 min.

37. The volume of brine in the tank after t min is V(t) = 100 + 2t gal, so the initial value
problem is
dx 3x
= 5− , x (0) = 50.
dt 100 + 2t

The integrating factor ρ (t ) = (100 + 2t)3/2 leads to the solution

50000
x(t ) = (100 + 2t ) − .
(100 + 2t )3/ 2

such that x(0) = 50. The tank is full after t = 150 min, at which time
x(150) = 393.75 lb.

38. (a) dx / dt = − x / 20 and x(0) = 50 so x(t ) = 50 e− t / 20 .

(b) The solution of the linear differential equation

dy 5x 5y 5 1
= − = e − t / 20 − y
dt 100 200 2 40

with y(0) = 50 is
y(t ) = 150 e − t / 40 − 100 e − t / 20 .

(c) The maximum value of y occurs when

y′(t ) = −
15 − t / 40
4
e
5
( )
+ 5e − t / 20 = − e − t / 40 3 − 4e − t / 40 = 0 .
4

We find that ymax = 56.25 lb when t = 40 ln(4/3) ≈ 11.51 min.

39. (a) The initial value problem

dx x
= − , x (0) = 100
dt 10

Section 1.5 6

for Tank 1 has solution x (t ) = 100 e − t /10 . Then the initial value problem

dy x y y
= − = 10 e − t /10 − , y( 0 ) = 0
dt 10 10 10

for Tank 2 has solution y (t ) = 10t e − t /10 .

(b) The maximum value of y occurs when

y′(t ) = 10 e − t /10 − t e − t /10 = 0

and thus when t = 10. We find that ymax = y(10) = 100e–1 ≈ 36.79 gal.

40. (b) 2 7
Assuming inductively that xn = t n e − t / 2 / n ! 2 n , the equation for xn+1 is

dxn +1 1 1 t n e−t / 2 1
= xn − xn +1 = − xn +1 .
dt 2 2 n ! 2 n +1 2

We easily solve this first–order equation with x n+1 ( 0) = 0 and find that

t n +1 e − t / 2
x n +1 = ,
(n + 1)! 2 n +1

thereby completing the proof by induction.

41. (a) A'(t) = 0.06A + 0.12S = 0.06 A + 3.6 e0.05t

(b) The solution with A(0) = 0 is

A(t) = 360(e0.06 t – e0.05 t),

so A(40) ≈ 1308.283 thousand dollars.

42. The mass of the hailstone at time t is m = ( 4 / 3)πr 3 = ( 4 / 3)πk 3t 3 . Then the equation
d(mv)/dt = mg simplifies to

tv' + 3v = gt.

The solution satisfying the initial condition v(0) = 0 is v(t) = gt/4, so v'(t) = g/4.

43. The solution of the initial value problem y ′ = x − y, y( −5) = y0 is

y( x ) = x − 1 + ( y0 + 6)e − x − 5 .

Section 1.5 7

Substituting x = 5, we therefore solve the equation 4 + ( y0 + 6)e −10 = y1
with y1 = 3.998, 3.999, 4, 4.001, 4.002 for the desired initial values
y0 = –50.0529, –28.0265, –6.0000, 16.0265, 38.0529, respectively.

44. The solution of the initial value problem y ′ = x + y, y( −5) = y0 is

y( x ) = − x − 1 + ( y0 − 4)e x + 5 .

Substituting x = 5, we therefore solve the equation −6 + ( y0 − 4)e10 = y1
with y1 = –10, –5, 0, 5, 10 for the desired initial values
y0 = 3.99982, 4.00005, 4.00027, 4.00050, 4.00073, respectively.

Section 1.5 8

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