6.1 slope fields (skip euler’s method) apply & practice 6.1 · ch 06 homework complete...

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Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson S. Stirling 2011-2012 Page 1 of 13 6.1 Slope Fields (skip Euler’s Method) Apply & Practice 6.1: P 391-294 #1, 3, 5, 10, 11, 13, 15, 19, 21, 23, 27, 31, 33 #37, 39, 43, 45, 49, 51, 53 – 56, 57 3. Differential equation: 2 2 2 xy y x y ′= Solution: 2 2 x y Cy + = : Differentiate: ( ) 2 2 2 2 2 2 2 2 x yy Cy x Cy yy x C yy x y C y + = = = ′= Check in diff. equ: 2 2 2 2 2 x xy C y x y = get left to match 2 2 2 xy Cy y = mult by y y replace Cy 2 2 2 2 2 xy x y y = + simplify 2 2 2 xy x y = checks! 5. Differential equation: 0 y y ′′ + = Solution: 1 2 cos sin y C x C x = + : Differentiate: 1 2 sin cos y C x C x ′=− + 1 2 cos sin y C x C x ′′ = − Check in diff. equ: 1 2 1 2 cos sin cos sin 0 C x C x C x C x + = 2 2 sin 0 C x = checks! Matches! See if (0) y gives you –2. Works! 11. 2 2 6 x y e = 4 y xy ′=− (0) 6 y = See if (0) y gives you 6. Works! Show that the derivative = 4 xy .

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Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 1 of 13

6.1 Slope Fields (skip Euler’s Method)

Apply & Practice 6.1: P 391-294 #1, 3, 5, 10, 11, 13, 15, 19, 21, 23, 27, 31, 33 #37, 39, 43, 45, 49, 51, 53 – 56, 57

3. Differential equation: 2 2

2xyyx y

′ =−

Solution: 2 2x y Cy+ = : Differentiate:

( )

2 22 22 2

22

x yy Cyx Cy yyx C y y

xyC y

′ ′+ =′ ′= −

′= −

′ =−

Check in diff. equ: 2 2

2 22x xy

C y x y=

− −

get left to match 2

22

xyCy y

=−

mult by yy

replace Cy 2 2 2

22

xyx y y

=+ −

simplify 2 2

2xyx y

=−

checks!

5. Differential equation: 0y y′′ + = Solution: 1 2cos siny C x C x= + : Differentiate: 1 2sin cosy C x C x′ = − + 1 2cos siny C x C x′′ = − − Check in diff. equ:

1 2 1 2cos sin cos sin 0C x C x C x C x− − + − = 22 sin 0C x− = checks!

Matches!

See if (0)y gives you –2. Works!

11. 226 xy e−=

4y xy′ = − (0) 6y =

See if (0)y gives you 6. Works!

Show that the derivative = 4xy− .

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 2 of 13

13. Find derivatives:

(4)

3cos3sin3cos

3sin3cos

y xy xy xy x

y x

=′ = −′′ = −′′′ =

=

.

Test:

( )

(4) 16 03cos 16 3cos 0

45cos 0

y yx x

x

− =

− =

− ≠

3cosy x= is NOT a solution.

15. Find derivatives: 2

2

2

2

(4) 2

24

816

x

x

x

x

x

y ey ey ey e

y e

=

′ = −

′′ =

′′′ = −

=

.

Test:

( )(4)

2 2

16 0

16 16 0

0 0

x x

y y

e e− −

− =

− =

=

2xy e−= IS a solution.

19. Given: 2y x= 2y x′ =

( ) ( )3

2 3

3

2

2 2

0

x

x

x

xy y x e

x x x x e

x e

′ − =

− =

Not a solution.

21. Given: ( )2 2 xy x e= +

( ) ( )2

2

2 2

4 2

x x

x x

y x e x e

y x xe x e

′ = + +

′ = + +

( ) ( )( )2 2 3

2 2 3 2 2 3

3 3

4 2 2 2

4 2 4 2

x x x x

x x x x

x x

x x xe x e x e x e

x x e x e x x e x ex e x e

+ + − + =

+ + − − =

=

Is a solution.

23. Given: lny x=

1y x′ =

( ) ( ) 31 2 ln xx x x ex − ≠

Not a solution.

→ 22 xy Ce−′ = −

Works!

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 3 of 13

33. Differential equation: 9 0y y′′ + = General Solution: 1 2sin3 cos3y C x C x= + : Differentiate:

1 2

1 2

1 2

sin3 cos33 cos3 3 sin3

9 sin3 9 cos3

y C x C xy C x C xy C x C x

= +′ = −′′ = − −

Check in diff. equ: ( ) ( )1 2 1 2

1 2 1 2

9 sin3 9 cos3 9 sin3 cos3 09 sin3 9 cos3 9 sin3 9 cos3 0

0 0

C x C x C x C xC x C x C x C x

− − + + =

− − + + ==

checks!

Initial conditions: 26

y π⎛ ⎞ =⎜ ⎟⎝ ⎠

, 16

y π⎛ ⎞′ =⎜ ⎟⎝ ⎠

Find 1C :

1 2

1 2

1 2

1 2

1

sin3 cos3

2 sin3 cos36 6

2 sin cos2 2

2 1 02

y C x C x

C C

C C

C CC

π π

π π

= +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +=

i i

Find 2C :

( )

1 2

2

2

2

1 3 cos3 3 sin6 6

1 3 2 cos 3 sin2 2

1 6 0 3 113

C C

C

C

C

π π

π π

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

− =

i i

Particular Solution: 12sin3 cos33

y x x= −

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 4 of 13

Oscillates. Same vertically & when x = 0, dy/dx = 1. When x = π, dy/dx = 1. Matches (b).

Oscillates. Same vertically & when x = 0, dy/dx = 0. Matches (c).

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 5 of 13

6.2 Differential Equations: Growth and Decay

Apply & Practice 6.2 Set 1: P 400-402 #1, 3, 6, 9, 11, 13, 15, 18, 19 Separation of variables. #31, 32 Think slope fields!

9. ( )21 2 0x y xy′+ − =

( )2

2

2

1 2 0

21

21

x y xy

dy xydx xdy x dxy x

′+ − =

=+

=+∫ ∫

21u x= + , 2 du x dx= , 2du dx

x=

2ln 2

x duyu x

= ∫ i

( )( )

( )

21

2

2

ln ln 1

ln ln 1 ln

ln ln 1

y x C

y x C

y C x

= + +

= + +

⎡ ⎤= +⎣ ⎦

( )

( )( )

( )

22ln 1ln

22

22

2

1

1

1

C xye e

y C x

y C x

y C x

⎡ ⎤+⎣ ⎦=

= +

= ± +

= +

6. 3

xyy

′ =

12

32 21

32 2

3

3

3 22 3

9 4

dy xdx y

y dy x dx

y x C

y x C

=

=

= +

= +

∫ ∫

3. 2dy ydx

= +

1

1

1

ln 2

21

2

2ln 2

22

y x C

C x

x

dy ydxdy dx

ydy dx

yy x C

e ey e ey Ce

+ +

+=

=+

=+

+ = +

=

+ = ±

= −

∫ ∫

i

1.

2

2

2

1 22

dy xdxdy x dx

y x x C

= +

= +

= + +

∫ ∫

11.

( )

2

2

1

1

dQ kdt tdQ k t dt

Q k t C

kQ Ct

=

=

= − +

= − +

∫ ∫

13.

( )

( )

( )

2

2

250

250

12502

5002

dN k sdsdN k s ds

N k s s C

kN s s C

= −

= −

⎛ ⎞= − +⎜ ⎟⎝ ⎠

= − − + +

∫ ∫

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 6 of 13

15b.

( )

( )

22

2

2

2

21

1ln 6 2

12

12

6

6

1ln 62

6

6

x Cy

xC

x

dy x ydx

dy x dxy

y x C

e e

y e e

y Ce

− +−

= −

=−

− − = +

=

− =

= +

∫ ∫

Initial Condition: ( )21 0

20 66

CeC

−= += −

Soln. 21

26 6x

y e−

= −

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 7 of 13

→ When 0y′ > .

Apply & Practice 6.2 Set 2: P 400-402 #23, 26 Thm 6.1 #33, 39, 43, 47, 57 Thm 6.1 apps.

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 8 of 13

$1822.12≈

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 9 of 13

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 10 of 13

6.3 Differential Equations: Separation of Variables

Apply & Practice 6.3: P 413 – 414 #1, 3, 5, 6, 7, 10, 14, 16, 18, 21, 23, 24, 45 – 47, 55 – 58

→ 22 xy Ce−′ = −

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 11 of 13

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 12 of 13

Ch 06 Homework Complete Solutions: S. Stirling Name _________________________________ Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2011-2012 Page 13 of 13