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SEC. 2.2 Separable Equations

In each of problems 1 through 8 solve the given differential equation :

ü 1. y '= x2y

„ SOL

„ y

„ x=

x2

y

, y ∫ 0

fl y „ y - x2

„ x = 0

fl y „ y - x2 „ x = 0

fl y

2

2-

x3

3= c, y ∫ 0

ü 2. y ' = x ^ 2y 1+ x3

„ SOL

„ y

„ x=

x2

y

1 + x3

, y ∫ 0

fl x

2

1 + x3 „ x - y „ y = 0

fl1

3ln 1 + x3 - y

2

2= c, y ∫ 0, x ∫ -1

ü 3. y '+ y2

sinx = 0

„ SOL

„ y

„ x= - y

2 sin xfl sin x „ x = - y-2 „ yfl -cos x = y-1 + c, " y ∫ 0; also y = 0 everywhere.

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ü 4. y '= 3 x2 - 1 3 + 2 y

„ SOL

„ y

„ x= 3 x

2 - 13 + 2 y , y ∫-3

2

fl 3 + 2 y „ y = 3 x2 - 1 „ xfl 3 y + y2 = x3 - x + c, y ∫

-3

2

ü 7. y '= x - „-x y + „y

„

SOL

„ y

„ x=

x - ‰- x y + ‰ y , y + ‰

y∫ 0

fl y + ‰ y „ y = x - ‰- x „ xfl

y2

2+ ‰

y=

x2

2+ ‰

- x+ c, y + ‰ y ∫ 0

ü 8. y '= x21 + y2

„ SOL

„ y

„ x=

x2

1 + y2fl x

2„ x = 1 + y2 „ y

fl x

3

3=

y

2+

y3

3+ c

In each of problems 9 through 20 :

(a) Find the solution of the given initial value problem in explicit form.

(b) Plot the graph of the solution.

(c) Determine (at least approximately) the interval in which the solution is defined.

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ü 9. y '= 1 - 2 x y2, y 0= -16

„ SOL

(a)

„ y

„ x= 1 - 2 x y2

fl y-2

„ y = 1 - 2 x „ xfl - y

-1= x - x

2+ c

Since y0 = -1 6fl 6 = c

fl y =1

x2 - x - 6, x2 - x - 6 ∫ 0

(b)

-4 -2 2 4 6 8

-0.6

-0.4

-0.2

0.2

0.4

0.6

(c)

x2

- x - 6 ∫ 0 ñ x - 3 x + 2 ∫ 0 ñ x ∫ 3 or x ∫ -2fl the solution is defined on - ¶, - 2 -2, 3 3, ¶.

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ü 10. y ' = 1- 2 xy , y 1=-2

„ SOL

(a)

„ y

„ x=

1 - 2 x y

fl y

2

2= x - x

2+ c

Since y1 = -2,fl c = 2

fl y = ≤ -2 x - 2 x + 1fl y1 = ≤ -2 -1 2 = ≤2, take y = - -2 x - 2 x + 1

(b)

-2 -1 1 2 3

-2.0

-1.5

-1.0

-0.5

(c)

Since - 2 x - 2 x + 1 ¥ 0fl x - 2 x + 1 § 0fl -1 § x § 2

fl the solution is defined on -1, 2.

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ü 11. x ‚ x + y „-x

‚ y = 0, y 0 = 1

„ SOL

(a)

x „ x + y ‰- x

„ y = 0

fl x ‰ x

„ x + y „ y = 0

fl x ‰ x

- ‰ x

+ y

2

2= c

Since y0 = 1,fl c = -

1

2

fl x ‰ x

- ‰ x

+ y2

2=

-1

2

fl y2

= -1 - 2 x ‰ x + 2 ‰ x

fl y = ≤ -1 - 2 x ‰ x + 2 ‰ x

fl y0 = ≤ -1 + 2 = ≤1, take y = -1 - 2 x ‰ x + 2 ‰ x

(b)

-2.0 -1.5 -1.0 -0.5 0.5 1.0

0.2

0.4

0.6

0.8

1.0

(c)

Since -1 - 2 x ‰ x + 2 ‰ x ¥ 0, and x = -1.67835, x = 0.768039 are the solutions of -1 - 2 x ‰ x + 2 ‰ x = 0

by Newton's method

fl the solution is defined on -1.67835, 0.768039.

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ü 12.‚ r ‚q = r 2q , r 1= 2

„ SOL

(a)

„ r

„ q =

r 2

q , q ∫ 0

fl q -1

„ q = r -2

„ r

fl ln q = - r -1 + cSince r 1 = 2fl c =

1

2

fl r -1

=1

2- ln q

fl r =2

1 - 2 ln q , 1 - 2 ln q ∫ 0

(b)

-4 -2 2 4

-10

-5

5

(c)

Since 1 - 2 ln q ∫ 0ñln q ∫ 1 2ñq ∫ ≤exp1 2 º ≤1.64872flthe solution is defined on - ¶, - exp 1 2 -exp1 2, 0 0, exp1 2 exp1 2, ¶.

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ü 13. y ' = 2 x y + x2 y , y 0 =-2

„ SOL

(a)

„ y

„ x=

2 x

y + x2 y

=2 x

y1 + x2fl

2 x

1 + x2„ x = y „ y

fl ln1 + x2 = y2

2+ c

Since y0 = -2fl c = -2fl ln1 + x2 = y

2

2- 2

fl y = ≤ 2 ln1 + x2 + 4 , since y0 = -2, take y = - 2 ln1 + x2 + 4

(b)

-6 -4 -2 2 4 6

-3.0

-2.8

-2.6

-2.4

-2.2

-2.0

(c)

Since 2 ln1 + x2 + 4 ¥ 0, " x œ flthe solution is defined on .

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ü 14. y ' = x y31 + x2

-12, y 0 = 1

„ SOL

(a)

„ y

„ x= x y

31 + x2-12

fl y-3

„ y = x1 + x2-12 „ xfl

-1

2 y

-2= 1 + x212 + c

Since y0 = 1fl c =

-3

2

fl y-2

= -2 1 + x212 + 3

fl y = ≤1

-2 1 + x212 + 3;

since y0 = 1, take y = 1-2 1 + x212 + 3

, - 2 1 + x212 + 3 > 0

(b)

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

1

2

3

4

(c)

Since -2 1 + x212 + 3 > 0ñ3 > 2 1 + x212ñ1 + x212 < 3 2ñ x2 < 5 4ñ- 5 4 < x < 5 4 º 1.11803flthe solution is defined on -1.11803, 1.11803.

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30. Consider the equation ‚ y ‚ x = y - 4 x x - y . (i)

ü (a) Show that it can be rewritten as

‚ y‚ x = y x - 4 1 - y x; (ii)

thus it is homogeneous.

„ pf

„ y

„ x=

y - 4 x

x - y=

x y x - 4 x1 - y x =

y x - 41 - y x , x ∫ 0

ü (b) Introduce a new dependent variable v so that v = y x, or y = x v x. Express ‚ y‚ x in terms of x, v, and ‚ v‚ x.

„ so l

Let v = y x ñ y = x v x, then„ y

„ x=

x v x„ x

= v x + x „ v„ x

ü (c) Replace y and ‚ y‚ x in ‚ y‚ x = y x - 4 1 - y x by the expressions from part (b) that involve v and ‚ v‚ x.Show that the resulting differential equation is

v + x ‚ v ‚ x = v - 4 1 - v or x ‚ v‚ x = v2 - 4 1 - v. (iii)

Observe that Eq.(iii) is separable.

„ so l

„ y

„ x=

y x - 41 - y x ñ v x + x

„ v

„ x=

v - 4

1 - vñ x

„ v

„ x=

v2 - 4

1 - v

ü (d) Solve Eq. (iii), obtaining v implicitly in terms of x.

„ so l

x„ v

„ x=

v2 - 4

1 - v

fl x-1

„ x =1 - v

v2 - 4

„ v

fl ln x = - 14

ln 2 - v - 34

ln 2 + v + c

fl ln x + 14

ln 2 - v + 34

ln 2 + v = cfl ln x2 - v14 2 + v34 = c1

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ü (e) Find the solution of Eq. (i) by replacing v by y x in the solution in part (d).

„ so l

flln

x

2-

v

14

2+

v

34

=

ln x

2-

y

x

14

2+

y

x

34

=

c1

fl x2 - y x14 2 + y x34 = c2fl x42 - y x 2 + y x3 = 2 x - y 2 x + y3 = cü (f) Draw a direction field and some integral curves for Eq. (i). Recall that the right side of Eq. (i) actually depends only on the

ratio y x. This means that integral curves have the same slope at all points on any given straight line through the origin,although the slope changes from one line to another. Therefore the direction filed and the integral curves are symmetric with

respect to the origin. Is this symmetry property evident from your plot?

„ direction field

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„ integral curves for Eq. (i)

Yes, the direction filed and the integral curves are symmetric with respect to the origin.

ü 31.‚ y ‚ x = x2 + xy + y2x2

„ SOL

(a) Show that the given equation is homogeneous.

„ y

„ x=

x2 + xy + y2

x2

= 1 + y

x

+ y

x

2

, x ∫ 0

(b) Solve the differential equation.

Let v = y x ñ y = x v x, thenv + x

„ v

„ x

= 1 + v + v2

fl x„ v

„ x= 1 + v2

fl x-1

„ x =1

1 + v2„ v

fl ln x = arctanv + c = arctan y x

+ c1

fl x = exparctan y x

+ c1

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(c) Draw a direction field and some integral curves. Are they symmetric with respect the

origin?

„ direction field

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„ integral curves


ü 32.‚ y ‚ x = x2 + 3 y22 x y

„ SOL


„ y

„ x=

x2 + 3 y2

2 x y=

x

2 y+

3 y

2 x=

1

2 y x +3

2

y

x

, x ∫ 0



„ v

„ x

=1

2 v

+3

2

v

fl x„ v

„ x=

1 + v2

2 v

fl x-1

„ x =2 v

1 + v2„ v

fl ln x = ln1 + v2 + c = ln 1 + y x

2

+ c1

fl ln x 1 + y2 x-2-1 = c1fl x 1 + y2 x-2

-1 = c

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origin?

„ direction field

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„ integral curves


ü 33.‚ y ‚ x = 4 y - 3 x 2 x - y

„ SOL


„ y

„ x=

4 y x - 32 - y x , x ∫ 0

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„ v

„ x

=4 v - 3

2 -v

fl x

„ v

„ x=

4 v - 32 - v - v =

v2 + 2 v - 3

2 - vfl x

-1„ x =

2 - v

v2 + 2 v - 3

„ v

fl ln x = 14

ln -3 -1 + v - 54

ln 3 3 + v + c1fl ln x4 = ln 3 v - 1 - ln 3 3 + v5 + c2fl ln

x

4

= ln 3 v - 1 - ln 3 3 +

v

5

+c2

fl ln x4 - ln 3 x-1 y - 1 + ln 3 3 + x-1 y5 = c2fl ln x4 3-1 x-1 y - 1-1 35 3 + x-1 y5 = c2fl ln34 y - x-1 3 x + y5 = c2fl 34 y - x-1 3 x + y5 = cfl 34 3 x + y5 = c y - x

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origin?

„ direction field

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„ integral curves


ü 34.‚ y ‚ x = -4 x + 3 y 2 x + y

„ SOL


„ y

„ x=

-4 x + 3 y2 x + y

=-4 + 3 y x

2 + y x , x ∫ 0

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„ v

„ x

=-4 + 3 v

2 + v

fl x„ v

„ x=

-4 + 3 v2 + v

- v =-4 - 5 v - v2

2 + v

fl x-1

„ x =2 + v

-4 - 5 v - v2„ v

fl ln x = - 23

ln -4 - v - 13

ln 1 + v + c1fl ln x3 = -ln-4 - v2 - ln 1 + v + c1fl

ln x

3

+

ln-

4-

v

2

+

ln 1+

v

=

c1

fl ln x3 -4 - v2 1 + v = c1fl ln x3 -4 - x-1 y2 1 + x-1 y = c1fl ln-4 x - y2 x + y = c1fl -4 x - y2 x + y = c

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origin?

„ direction field

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„ integral curves


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