## anonymous one year ago find D.E x^2/a^2+y^2/b^2=1 ans. xyy''+xy'^2=yy'

1. anonymous

diff. twice and eliminate a^2 and b^2 from three equations.

2. anonymous

how @surjithayer

3. anonymous

4. ganeshie8

Have you used the surjithayer's hint and differentiated twice ?

5. anonymous

yeah

6. ganeshie8

7. anonymous

wait

8. anonymous

9. anonymous

10. anonymous

number 14

11. ganeshie8

whats the derivative of $$\dfrac{x^2}{a^2}$$ with respect to $$x$$ ?

12. anonymous

but i think , may solution is not right

13. anonymous

yes

14. ganeshie8

Keep in mind, $$a$$ and $$b$$ are constants here

15. ganeshie8

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}$ differentiating implicitly with respect to $$x$$ gives $\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0\tag{2}$ yes ?

16. anonymous

okay :)

17. anonymous

but , i did it first , but my answer was not right

18. ganeshie8

your work in that picture is wrong you're not treating a^2 as constant, instead trying to differentiate it and writing 2a.. which is wrong

19. anonymous

20. ganeshie8

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}$ differentiating implicitly with respect to $$x$$ gives $\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'}$ Plug this in first equation and get $\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1$ $\implies -xyy' +y^2 = b^2 \tag{2}$ differentiate this again and you're done!

21. anonymous

@ganeshie8 gets :)) thank you very much :))

22. ganeshie8

yw, are you able to differentiate it again and arrive at the answer ?

23. anonymous

@ganeshie8 im very confused

24. ganeshie8

remember product rule ?

25. anonymous

yeah i remember

26. anonymous

27. ganeshie8

b^2 is just a constant, like, 3^2 or 5^2

28. ganeshie8

it doesn't change, so the derivative is 0

29. anonymous

$-x y y'' . -1 + x y^2.y'=b^2$

30. ganeshie8

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}$ differentiating implicitly with respect to $$x$$ gives $\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'}$ Plug this in first equation and get $\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1$ $\implies -xyy' +y^2 = b^2 \tag{2}$ differentiating w.r.t $$x$$ again $-(yy' +y'xy' + y''xy) + 2yy'= 0$ $yy' = xyy'' + x{y'}^2$

31. ganeshie8

Recall below product rules $\left(f(x)*g(x)\right)' = f'(x)g(x) + f(x)g'(x)$ $\left(f(x)*g(x)*h(x)\right)' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)$

32. anonymous

@ganeshie8 i got the answer :))))

33. anonymous

34. ganeshie8

Give it a try first

35. anonymous

36. anonymous

@ganeshie8 number 11

37. ganeshie8

Looks you forgot to use product rule here ? |dw:1436688898948:dw|

38. UsukiDoll

isn't that we have to find the second and third derivatives and then plug them back into the equation to check if the equation we are given (that ans. thing) holds true?

39. anonymous

@ganeshie8

40. UsukiDoll

x and y's are variables treat a b c as constants which when the derivative is taken it goes to 0

41. anonymous

ai derivative of x is 1 = a (1) ?

42. ganeshie8

@UsukiDoll this is like constructing the differential equation from the given general solution by eliminating the arbitrary constants

43. UsukiDoll

oh I see so we're not verifying... we're finding a DE

44. ganeshie8

yeah we can think of the equations of given ellipses/parabolas as general solutions

45. anonymous

yeah , were finding D.E

46. ganeshie8

@jacalneaila did you find ur mistake in #11 ?

47. anonymous

yeah, i try to solve ut again , but may answer is different now :(

48. anonymous

*it

49. ganeshie8

show ur work

50. anonymous

wait , i dont have a cellphone

51. ganeshie8

#11 should be fairly simple compared to the earlier one

52. anonymous

$ax+yy'+a(1)?$

53. anonymous

@ganeshie8

54. ganeshie8

|dw:1436690038639:dw|

55. anonymous

ugh

56. anonymous

im still doing it , and still different

57. anonymous

@ganeshie8 Thank you very much , its very helpful :)) .

58. anonymous

@ganeshie8 Thank you again ^_^

59. UsukiDoll

so for this part axy to make it less confusing I would put a () like this a(xy) and then take the product rule of xy a(xy'+y) then I have to take the product rule for xy' a(xy''+y'+y') and then another product rule for xy'' a(xy'''+y''+y''+y'') so that leaves (xy'''+3y'') I think I'm missing a step.. that a is supposed to go away x.x

60. ganeshie8

the missing part is $$=0$$ in every line :)

61. UsukiDoll

oh boy xD! I was doing this while dealing with a migraine.

62. ganeshie8

From personal experience, regular diet+good sleep are the best medicine for migraine

63. UsukiDoll

yeah...that's what I lot of people say...but I'm such a nighthawk and it runs in the family.

64. ganeshie8

Also there are some ellergetic foods which you need to avoid eating

65. UsukiDoll

if it's chocolate, then I can't =(

66. ganeshie8

For me they are potato and some other vegetable whose name idk in english

67. ganeshie8

they depend on the person i think..

68. ganeshie8