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Arisa [49]
3 years ago
8

(X^2+y^2+x)dx+xydy=0 Solve for general solution

Mathematics
1 answer:
aksik [14]3 years ago
5 0

Check if the equation is exact, which happens for ODEs of the form

M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0

if \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.

We have

M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y

N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y

so the ODE is not quite exact, but we can find an integrating factor \mu(x,y) so that

\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0

<em>is</em> exact, which would require

\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}

\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}

Notice that

\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y

is independent of <em>x</em>, and dividing this by N(x,y)=xy gives an expression independent of <em>y</em>. If we assume \mu=\mu(x) is a function of <em>x</em> alone, then \frac{\partial\mu}{\partial y}=0, and the partial differential equation above gives

-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}

which is separable and we can solve for \mu easily.

-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}

\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x

\ln|\mu|=\ln|x|

\implies \mu=x

So, multiply the original ODE by <em>x</em> on both sides:

(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0

Now

\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy

\dfrac{\partial(x^2y)}{\partial x}=2xy

so the modified ODE is exact.

Now we look for a solution of the form F(x,y)=C, with differential

\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

The solution <em>F</em> satisfies

\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2

\dfrac{\partial F}{\partial y}=x^2y

Integrating both sides of the first equation with respect to <em>x</em> gives

F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)

Differentiating both sides with respect to <em>y</em> gives

\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y

\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C

So the solution to the ODE is

F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C

\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}

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aalyn [17]
64. Susan tipped at the rate of 75 dollars to 5 waiters. Question: She can afford to pay up to 90 dollars, How many waiters can she tipped if that is the case. Let’s solve first and identify how much will she be giving to each waitress with 75 dollars is to 5 waiters rate: => 75 dollars / 5 waiters = 15 dollars each waiter Now, she have 90 dollars => 90 – 75 = 15 Thus => 90 / 15 = 6 waiters she will be tipping.
5 0
3 years ago
PLEASE HELP
Tems11 [23]

Answer:

1.

5

x

−

2

y

=

4

; (−1, 1)

2.

3

x

−

4

y

=

10

; (2, −1)

3.

−

3

x

+

y

=

−

6

; (4, 6)

4.

−

8

x

−

y

=

24

; (−2, −3)

5.

−

x

+

y

=

−

7

; (5, −2)

6.

9

x

−

3

y

=

6

; (0, −2)

7.

1

2

x

+

1

3

y

=

−

1

6

; (1, −2)

8.

3

4

x

−

1

2

y

=

−

1

; (2, 1)

9.

4

x

−

3

y

=

1

;

(

1

2

,

1

3

)

10.

−

10

x

+

2

y

=

−

9

5

;

(

1

5

,

1

10

)

11.

y

=

1

3

x

+

3

; (6, 3)

12.

y

=

−

4

x

+

1

; (−2, 9)

13.

y

=

2

3

x

−

3

; (0, −3)

14.

y

=

−

5

8

x

+

1

; (8, −5)

15.

y

=

−

1

2

x

+

3

4

;

(

−

1

2

,

1

)

16.

y

=

−

1

3

x

−

1

2

;

(

1

2

,

−

2

3

)

17.

y

=

2

; (−3, 2)

18.

y

=

4

; (4, −4)

19.

x

=

3

; (3, −3)

20.

x

=

0

; (1, 0)

Find the ordered pair solutions given the set of x-values.

21.

y

=

−

2

x

+

4

; {−2, 0, 2}

22.

y

=

1

2

x

−

3

; {−4, 0, 4}

23.

y

=

−

3

4

x

+

1

2

; {−2, 0, 2}

24.

y

=

−

3

x

+

1

; {−1/2, 0, 1/2}

25.

y

=

−

4

; {−3, 0, 3}

26.

y

=

1

2

x

+

3

4

; {−1/4, 0, 1/4}

27.

2

x

−

3

y

=

1

; {0, 1, 2}

28.

3

x

−

5

y

=

−

15

; {−5, 0, 5}

29.

–

x

+

y

=

3

; {−5, −1, 0}

30.

1

2

x

−

1

3

y

=

−

4

; {−4, −2, 0}

31.

3

5

x

+

1

10

y

=

2

; {−15, −10, −5}

32.

x

−

y

=

0

; {10, 20, 30}

Find the ordered pair solutions, given the set of y-values.

33.

y

=

1

2

x

−

1

; {−5, 0, 5}

34.

y

=

−

3

4

x

+

2

; {0, 2, 4}

35.

3

x

−

2

y

=

6

; {−3, −1, 0}

36.

−

x

+

3

y

=

4

; {−4, −2, 0}

37.

1

3

x

−

1

2

y

=

−

4

; {−1, 0, 1}

38.

3

5

x

+

1

10

y

=

2

; {−20, −10, −5}

Part B: Graphing Lines

Given the set of x-values {−2, −1, 0, 1, 2}, find the corresponding y-values and graph them.

39.

y

=

x

+

1

40.

y

=

−

x

+

1

41.

y

=

2

x

−

1

42.

y

=

−

3

x

+

2

43.

y

=

5

x

−

10

44.

5

x

+

y

=

15

45.

3

x

−

y

=

9

46.

6

x

−

3

y

=

9

47.

y

=

−

5

48.

y

=

3

Find at least five ordered pair solutions and graph.

49.

y

=

2

x

−

1

50.

y

=

−

5

x

+

3

51.

y

=

−

4

x

+

2

52.

y

=

10

x

−

20

53.

y

=

−

1

2

x

+

2

54.

y

=

1

3

x

−

1

55.

y

=

2

3

x

−

6

56.

y

=

−

2

3

x

+

2

57.

y

=

x

58.

y

=

−

x

59.

−

2

x

+

5

y

=

−

15

60.

x

+

5

y

=

5

61.

6

x

−

y

=

2

62.

4

x

+

y

=

12

63.

−

x

+

5

y

=

0

64.

x

+

2

y

=

0

65.

1

10

x

−

y

=

3

66.

3

2

x

+

5

y

=

30

Part C: Horizontal and Vertical Lines

Find at least five ordered pair solutions and graph them.

67.

y

=

4

68.

y

=

−

10

69.

x

=

4

70.

x

=

−

1

71.

y

=

0

72.

x

=

0

73.

y

=

3

4

74.

x

=

−

5

4

75. Graph the lines

y

=

−

4

and

x

=

2

on the same set of axes. Where do they intersect?

76. Graph the lines

y

=

5

and

x

=

−

5

on the same set of axes. Where do they intersect?

77. What is the equation that describes the x-axis?

78. What is the equation that describes the y-axis?

Part D: Mixed Practice

Graph by plotting points.

79.

y

=

−

3

5

x

+

6

80.

y

=

3

5

x

−

3

81.

y

=

−

3

82.

x

=

−

5

83.

3

x

−

2

y

=

6

84.

−

2

x

+

3

y

=

−

12

Step-by-step explanation:

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3 years ago
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Serga [27]

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Total amount of Brian paid is $48.30.

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6 0
3 years ago
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Rewrite 19/19 as a whole number
Elis [28]

The answer is 1.

Because if you simplify (divide) 19/19 you get one.

4/4 would also be 1.  Any number divided to itself will always be 1.

Whereas 30/15 would be 2 because thirty divided by fifteen is 2

5 0
3 years ago
Martine wants to open an after-school tutoring service for students in 6th through 9th grade. How can a thorough knowledge of cu
grin007 [14]

Answer:

it would help her know how to prepare her teaching to match the students learning and expectations

Step-by-step explanation:

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Knowledge of these expectations would help to set Martine on the path of tutoring success and this would attract more students. So for her to have a strong tutoring business she has to know the approaches to use to make students strong academically, and how to match learning ability with her teaching.

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3 years ago
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