## Astrophysics one year ago @ganeshie8

1. Astrophysics

How can I tell where this solution is valid, the $(2x-y)dx+(2y-x)dy = 0$ solving this exact equation I got $\psi (x,y) = x^2-xy+y^2-7$ but I'm not sure how to tell where this is valid

2. ganeshie8

what are the initial conditions ?

3. Astrophysics

Oh oops! y(1) = 3

4. Mimi_x3

are you taking a DE COURSE???

5. Mimi_x3

study with me pls <3

6. ganeshie8

so the solution curve is $$x^2-xy+y^2-7=0$$

7. Astrophysics

Haha yeah I am

8. Astrophysics

Yes

9. Astrophysics

the back of the book says $|x|<\sqrt{\frac{ 28 }{ 3 }}$

10. Mimi_x3

STUDY WITH ME IM LIKE 5 weeks behind LOL in class we're like up to PDE's and im still stuggling with fourier :/

11. ganeshie8

|dw:1443339556634:dw|

12. ganeshie8

Notice that the solution curve is trapped between those vertical tangents we may use that observation to find the x bounds

13. Astrophysics

Yup I see, but I'm not sure how to do it algebraically

14. ganeshie8

As a start, implicitly differentiate the equation of solution curve and find an expression for $$y'$$

15. Astrophysics

kk

16. ganeshie8

alternatively you may think of the solution curve as a quadratic in $$y$$, then use the discriminant to see for what values of $$x$$, the value of $$y$$ is real

17. Astrophysics

$y' = \frac{ 2x-y }{ x-2y }$

18. Astrophysics

Ooh so how would you set that up using quadratic formula

19. ganeshie8

in above expression for derivative, $$x=2y$$ makes the derivative go wild, so vertical tangents must occur when $$x=2y$$ using that eliminate $$y$$ in the solution curve and solve $$x$$

20. Astrophysics

That works!

21. ganeshie8

If that derivative thing doesn't strike immediately, you may simply use the disrcriminat : $$\color{red}{y^2}-\color{red}{y}x+x^2-7=0$$ this is a quadratic in $$\color{red}{y}$$ : $$a=1$$ $$b=-x$$ $$c=x^2-7$$ Discriminant must be nonnegative for $$y$$ to be real. so $$x^2-4(1)(x^2-7)\ge 0$$, we can solve $$x$$

22. Astrophysics

Ha, I like the derivative thing, so this is finding the same thing as interval of definition?

23. ganeshie8

Yes this is like finding the largest interval in which the solution curve is defined

24. Astrophysics

So how come it doesn't say $-\sqrt{\frac{ 28 }{ 3 }} < x < \sqrt{\frac{ 28 }{ 3 }}$

25. Astrophysics

< = to I believe it should be

26. ganeshie8

dy/dx is not defined at either ends, so we must exclude them

27. Astrophysics

Oh right!

28. Astrophysics

Haha

29. ganeshie8

the derivative thingy is easy because the derivative is readily available for us : $$(2x-y)dx+(2y-x)dy = 0 \implies \dfrac{dy}{dx}=\dfrac{y-2x}{2y-x}$$ Clearly you don't wanto plug $$2y=x$$ in above equation

30. Astrophysics

Yup, I really like this method

31. Astrophysics

Solving O.D.E's isn't too bad, it's this interval stuff that can be confusing as it requires you to remember stuff from calc 1 >.<

32. Astrophysics

Thanks again

33. ganeshie8

the interval stuff is really not that easy, there are several things that you need to consider when working the domain of the solution

34. ganeshie8

this pdf talks about "why" domain is important and "how" to find the domain http://apcentral.collegeboard.com/apc/members/repository/ap07_calculus_DE_domain_fin.pdf

35. Astrophysics

Awesome, thanks by the way...what exactly is the difference from the problems that say "determine the interval in which the solution is defined" and from this where it asks you for "where solution is valid"

36. ganeshie8

both are same

37. Astrophysics

Ah ok haha, thanks xD

38. ganeshie8

except for philosophical differences, both refer to the same thing

39. Astrophysics

Ok cool, yeah I guess it's just they word each question differently, so I thought there must be something different about them even though they sound the same

40. ganeshie8

1) "determine the interval in which the solution is defined" 2) "where solution is valid" first one is in ur control, "You" can define the interval where your solution would be valid while modeling the differential equaiton based on physical constraints. for the second one, you need to work out the domain by looking at the solution but there is not much difference between them, at least i never worried about the differences..

41. Astrophysics

Ok that makes sense, like I had a separable equation problem earlier and it asked for where it was defined $\frac{ dy }{ dx } = \frac{ arcsinx }{ y^2(1-x^2)^{1/2} }$ I sort of just looked at the derivative here and got -1<x<1, but I guess I should be paying attention to vertical tangents as well?

42. ganeshie8

what are the initial conditions ?

43. Astrophysics

y(0)=1

44. ganeshie8

whats the solution ?

45. IrishBoy123

.

46. ganeshie8

you must consider the DE, the initial conditions and the solution while working the domain

47. ganeshie8

All 3 things can change the domain

48. Astrophysics

Ok yeah that's what I did, that was sort of a bad example

49. Astrophysics

I didn't include all the info

50. Astrophysics

The solution was $y(x) = \left[ \frac{ 3 }{ 2 }\arcsin^2x \right]^{1/3}+1$

51. Astrophysics

$y^2dy = \frac{ arcsinx }{ (1-x^2) }dx$ y(0)=1 so not too hard of a problem

52. ganeshie8

-1<x<1 looks good to me why do you think it is wrong ?

53. Astrophysics

Oh no I didn't think it was wrong it made sense to me, but I didn't really consider the vertical tangent

54. Astrophysics

Ok nvm, I think I have got a decent idea of this stuff now hehe, thanks!

55. ganeshie8

try finding the domain of this IVP $\dfrac{dy}{dx}=\dfrac{2}{\sqrt{y}};~~y(0) = 9$

56. Astrophysics

Ok give me a sec

57. Astrophysics

Omg C seems ugly X_X

58. Astrophysics

nvm its 18?

59. ganeshie8

you may simply use wolfram to find the solution http://www.wolframalpha.com/input/?i=solve+dy%2Fdx%3D2%2Fsqrt%28y%29%2C+y%280%29%3D9

60. Astrophysics

Ok haha, $y=3^{2/3}(x+9)^{2/3}$ so the domain then is x>-9

61. Astrophysics

$x \ge -9$

62. ganeshie8

why

63. ganeshie8

what happens if i plugin x = -10

64. Astrophysics

you get -3^(2/3)

65. Astrophysics

So that works

66. ganeshie8

so nothing wrong with x<-9 ?

67. Astrophysics

Nope, so all real numbers then?

68. ganeshie8

Look at the graph of solution again : |dw:1443344660260:dw|

69. ganeshie8

then look at the differential equation again

70. ganeshie8

the graph is not matching with the given differential equation, why ?

71. ganeshie8

|dw:1443344789571:dw|

72. Astrophysics

Nooo, because we then get undefined

73. ganeshie8

right, the slope of tangent to solution curve, y, is negative for y < -9, yes ?

74. Astrophysics

Yup

75. ganeshie8

|dw:1443345018801:dw|

76. Astrophysics

But see that's where I was thinking, when we look at the derivative itself we would have to consider the vertical tangets

77. Astrophysics

tangets*

78. ganeshie8

its not primarily about vertical tangents here its much more deeper than that, the slope field of the given differential equation cannot have negative slope segments

79. ganeshie8

so we must exclude x<-9 from the domain of the solutions

80. Astrophysics

Ah!

81. ganeshie8

as you can see, finding domain is not so straightforward and definitely not easy

82. Astrophysics

Yeah, it's pretty confusing haha

83. Astrophysics

So we're sort of restricting it?

84. ganeshie8

since the derivative is always "positive", x <= -9 makes no sense, so the domain is x > -9

85. Astrophysics

Ok I see

86. Astrophysics

Oh haha I'm reading the link, that's where you got the problem

87. Astrophysics

This stuff is confusing haha it's like I get it and at the same time I don't

88. ganeshie8

domains for differential equations are not same as that of algebraic equations nothing to feel bad about, it takes a while to get used to...

89. Astrophysics

I guess, I'll have to do lots of practice

90. Astrophysics

Well thanks a lot ganeshie it's nearly 4 am, been doing this stuff nearly all day haha, time for sleep, take care, bye!

91. ganeshie8

tc, have good sleep :)