## anonymous one year ago don't know how to solve this differential equation

1. anonymous

2. freckles

$\frac{dy}{y}= 6x \sqrt{3x^2-1} dx$

3. freckles

integrate both sides

4. anonymous

okay real quick wouldn't it be dy/dx on the left side?

5. UsukiDoll

separation of variables.... in the form of h(y) dy = f(x) dx

6. freckles

i multiplied both sides by dx

7. UsukiDoll

you need all y's to the left and all x's to the right... then integrate both sides..

8. freckles

i also divided both sides by y and multiplied both sides by 6x

9. anonymous

okay i see one sec let me integrate

10. anonymous

okay so im getting $\frac{ 2(3x^2 -1)^3/2 }{ 3}+C$

11. anonymous

not sure how to do the left side

12. freckles

$\frac{2}{3}(3x^2-1)^\frac{3}{2}+C \text{ is \right for the \right hand side}$

13. freckles

the left hand hand is just ln|y|+k but you only need one constant of integration in your equation

14. freckles

$\ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K$

15. anonymous

wait how did you get that from dy/y?

16. freckles

$\frac{1}{6x} \frac{dy}{dx}=y \sqrt{3x^2-1} \\ \text{ divide both sides by } y \\ \frac{1}{6x} \frac{1}{y} \frac{dy}{dx} = \sqrt{3x^2-1} \\ \text{ multiply both sides by } 6x \\ \frac{1}{y} \frac{dy}{dx}= 6x \sqrt{3x^2-1} \\ \text{ multiply both sides by } dx \\ \frac{1}{y} dy= 6x \sqrt{3x^2-1} dx \\ \text{ or this can also be written as } \frac{dy}{y}=6x \sqrt{3x^2-1} dx$

17. UsukiDoll

1/y dy the antiderivative of 1/y is just ln y

18. freckles

oh not how I got dy/y

19. freckles

but how you integrate it

20. UsukiDoll

but we're still not done because we need a y by itself... we have to take e^ {} on everything so that we have y = (all the contents)

21. freckles

the derivative of ln|y| w.r.t to y is y'/y=1/y

22. anonymous

no i ment how did you integrate dy/y

23. freckles

$\int\limits \frac{du}{u}=\ln|u|+K \text{ since } \frac{d}{du}(\ln|u|+K)=\frac{1}{u}+0$

24. UsukiDoll

first rewrite this as 1/y dy then the antiderivative of 1/y is ln y it's a logarithm the derivative is 1/y (1) because we also have to take the derivative of the denominator but it's just 1, so it's 1/y only

25. freckles

well actually it is a definition in most calculus books the following: $\ln(x)=\int\limits_1^x \frac{1}{t} dt$

26. anonymous

okay i get it so now i just isolate y right?

27. freckles

you can if you are asked to find an explicit form for y

28. anonymous

it just says solve the differential equation ... so i assume i just solve for y

29. freckles

you could stop at $\ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K$ then but solving this one for y is doable

30. UsukiDoll

e^{ln y} e and ln are inverses of each other so it cancels out for y we're you given initial conditions in your problem ? if you were, then we can solve... if not we can just take e^{} on both sides

31. anonymous

okay but then how do i get rid of the absolute value ?

32. UsukiDoll

$\large \ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K$ $\large e^{ln|y|}=e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K}$ $\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K}$

33. freckles

$e^{\ln|y|}=|y|$

34. anonymous

these are my answer choices by the way and none of the above

35. freckles

as long as y is not 0 of course

36. freckles

e^K=C @Jdosio

37. anonymous

38. UsukiDoll

forgot that part... e^K = C substitution.

39. freckles

$\text{ assuming } y>0 \text{ we do have } \\ y=e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K} \\ \\ \text{ but we can also rewrite this } \\ y=e^{k} e^{\frac{2}{3} (3x^2-1)^\frac{3}{2}} \text{ by law of exponents } \\ \text{ so now \it doesn't matter much about the } y>0 \text{ since } C \\ \text{ could be negative or positive } \\ \\ y=C e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}$

40. UsukiDoll

$\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}e^{K}$ $\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}C$

41. freckles

@UsukiDoll is not really part you forgot. it is just a little something extra that isn't really needed but people write because it does look prettier

42. anonymous

okay so wtf is K then

43. freckles

e^k is a constant

44. freckles

C is a constant

45. freckles

you are still replacing an unknown constant with another unknown constant

46. anonymous

why did you use K instead of c then

47. freckles

because you can

48. freckles

you can mean any letter to be a constant

49. freckles

I used K as a constant above we ended up with e^K which still an unknown constant you can replace that with another unknown constant call it C

50. UsukiDoll

you can have any letters as a constant tbh. but it's usually C

51. anonymous

okay i understand that the e^anyunknown constant equals another constant my question was why you used two different symbols for constant in the same equation

52. freckles

e^K =C doesn't imply K=C

53. UsukiDoll

I usually just leave the solution alone...unless it begs to be changed .

54. anonymous

okay so then its just to keep track of which constant is which , okay that sort of makes sense

55. freckles

it seems funny to me to replace e^C with C

56. freckles

it is like replace e^5 with 5

57. freckles

even though e^5 doesn't equal 5

58. anonymous

i find it soothing to just be able to take out constants like that

59. freckles

but yeah you could have just said e^C then replace it with C i think some people do that

60. freckles

take out constants?

61. anonymous

combine

62. anonymous

anyway thank you very much for the help guys i wish i could give you both a medal :D

63. anonymous

ugh who do i give it too D:

64. UsukiDoll

I gave one to freckles already... so......?