anonymous one year ago ODE y^2=cx+1/8 c^3 ------ y=2xy'+y^2(y')^3

1. UsukiDoll

??? $y^2=cx+\frac{1}{8}c^3..............y=2xy'+y^2(y')^3$ or $y^2=cx+\frac{1}{8c^3}..............y=2xy'+y^2(y')^3$

2. UsukiDoll

I don't know what's going on.. can you go into more detail?

3. anonymous

$y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3$

4. anonymous

y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3 if the given left is a solution od the ODE right :))

5. UsukiDoll

so we are differentiating? one part of the equation with the fraction is still confusing.

6. UsukiDoll

so in the end we should have something like $y=2xy'+y^2(y')^3$?

7. anonymous

yes

8. UsukiDoll

could you rewrite the first equation.. it's the fraction portion that is making me lost. I don't know if $c^3y$ is in the denominator or the numerator

9. anonymous

its only c^3

10. anonymous

1/8 c^3

11. UsukiDoll

-_-! $y^2=cx+\frac{1}{8}c^3.$

12. UsukiDoll

omg I apologize the latex is crazy |dw:1436693748089:dw|

13. UsukiDoll

so we let the x and y be the variables and we treat c as a constant.

14. anonymous

yeah

15. anonymous

sorry

16. anonymous

@ganeshie8

17. UsukiDoll

well on the farthest term the one with the 1/8c^3 if c is a constant and if all numbers are constants.. we will just have 0

18. anonymous

i dont know how to solve it :(

19. UsukiDoll

do you know the derivative of y^2 ?

20. anonymous

2yy'

21. UsukiDoll

hmm... I would just leave it as 2y unless we are using implicit differentiation as well?

22. UsukiDoll

what about the derivative of cx?

23. anonymous

c(1) >

24. UsukiDoll

let's see... product rule is f(x)g'(x)+f'(x)g(x) so we let f(x) = c and g(x) = x what is the derivative of f'(x) and what is the derivative of g'(x) ?

25. anonymous

c (1) + cx ?

26. UsukiDoll

well .. the derivative of g(x) is 1 yes.. c(1) + that part is correct so for f'(x)g(x) f(x) = c and g(x) = x so we leave g(x) alone and take the derivative of f(x) = c. Since c is a constant, the derivative is 0 so we have c(1) +x(0) c+0 c

27. anonymous

$2yy'= c+1/8 c^3$

28. UsukiDoll

why is 1/8c^3 still there? a number like 1/8 is just a constant and c^3 is a constant too? I'm lost as the hills XD

29. anonymous

sorry

30. anonymous

XD

31. anonymous

$3/8 c^2?$

32. UsukiDoll

uh ._. I have no idea what's going on now *mind blown*

33. anonymous

1/8c^3 is not a constant

34. UsukiDoll

1/8 is a number... numbers are constants... derivatives of any number go to 0.

35. anonymous

what is the next step

36. UsukiDoll

hmmm is c also treated as a variable? I can't even...

37. UsukiDoll

usually x and y are variables and anything but those letters are treated as a constant...

38. anonymous

3/8c^2 ?

39. anonymous

uh ._. I have no idea what's going on now *mind blown* HAHA

40. UsukiDoll

in the end we are supposed to have x's and y's somehow.. so that c has to be treated as a constant.

41. anonymous

pass muna nahihirapan na ako e ahahaha

42. anonymous

i dont understand

43. UsukiDoll

like... maybe factor one c out to make it like $y^2 = c(x+\frac{1}{8}c^2)$

44. UsukiDoll

then differentiate...

45. UsukiDoll

like we know that the derivative of $y^2 \rightarrow 2y y'$ and then product rule for c()

46. anonymous

what is mean of ODE ??

47. UsukiDoll

ODE means an ordinary differential equation...

48. UsukiDoll

$y^2 dy = c(x+\frac{1}{8}c^2) dx$ if it's a separable equation, then we have to integrate both sides.

49. anonymous

what is the next step

50. UsukiDoll

integrate both sides.

51. anonymous

oh!! so we are given $y=2xy' +y^{2}(y')^{3} and y ^{2}=cx+1\div8c ^{3} is \it ???$

52. UsukiDoll

@sohailiftikhar supposedly we are differentiating $y^2=cx+\frac{1}{8}c^3.$ and by doing that we get to $y=2xy'+y^2(y')^3$

53. anonymous

just put value of $y ^{2}$ in equation of y then take derivative of it ...

54. anonymous

is that ??

55. UsukiDoll

this isn't even my question, so .. yeah

56. anonymous

how it can be ...?

57. UsukiDoll

and with vague instructions, I can't go further... it's like hello I'm not a mind reader like what does the question want really?

58. anonymous

I think we are give two equations and we have to find dy/dx

59. UsukiDoll

this is similar to OP's previous question though.

60. anonymous

I can't get it what is that .....sorry .

61. UsukiDoll

I don't get it either.. let's get out of here.

62. anonymous

yes..