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anonymous

  • one year ago

don't know how to solve this differential equation

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  1. anonymous
    • one year ago
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  2. freckles
    • one year ago
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    \[\frac{dy}{y}= 6x \sqrt{3x^2-1} dx\]

  3. freckles
    • one year ago
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    integrate both sides

  4. anonymous
    • one year ago
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    okay real quick wouldn't it be dy/dx on the left side?

  5. UsukiDoll
    • one year ago
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    separation of variables.... in the form of h(y) dy = f(x) dx

  6. freckles
    • one year ago
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    i multiplied both sides by dx

  7. UsukiDoll
    • one year ago
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    you need all y's to the left and all x's to the right... then integrate both sides..

  8. freckles
    • one year ago
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    i also divided both sides by y and multiplied both sides by 6x

  9. anonymous
    • one year ago
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    okay i see one sec let me integrate

  10. anonymous
    • one year ago
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    okay so im getting \[\frac{ 2(3x^2 -1)^3/2 }{ 3}+C\]

  11. anonymous
    • one year ago
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    not sure how to do the left side

  12. freckles
    • one year ago
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    \[\frac{2}{3}(3x^2-1)^\frac{3}{2}+C \text{ is \right for the \right hand side}\]

  13. freckles
    • one year ago
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    the left hand hand is just ln|y|+k but you only need one constant of integration in your equation

  14. freckles
    • one year ago
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    \[\ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K\]

  15. anonymous
    • one year ago
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    wait how did you get that from dy/y?

  16. freckles
    • one year ago
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    \[\frac{1}{6x} \frac{dy}{dx}=y \sqrt{3x^2-1} \\ \text{ divide both sides by } y \\ \frac{1}{6x} \frac{1}{y} \frac{dy}{dx} = \sqrt{3x^2-1} \\ \text{ multiply both sides by } 6x \\ \frac{1}{y} \frac{dy}{dx}= 6x \sqrt{3x^2-1} \\ \text{ multiply both sides by } dx \\ \frac{1}{y} dy= 6x \sqrt{3x^2-1} dx \\ \text{ or this can also be written as } \frac{dy}{y}=6x \sqrt{3x^2-1} dx\]

  17. UsukiDoll
    • one year ago
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    1/y dy the antiderivative of 1/y is just ln y

  18. freckles
    • one year ago
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    oh not how I got dy/y

  19. freckles
    • one year ago
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    but how you integrate it

  20. UsukiDoll
    • one year ago
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    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
    • one year ago
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    the derivative of ln|y| w.r.t to y is y'/y=1/y

  22. anonymous
    • one year ago
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    no i ment how did you integrate dy/y

  23. freckles
    • one year ago
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    \[\int\limits \frac{du}{u}=\ln|u|+K \text{ since } \frac{d}{du}(\ln|u|+K)=\frac{1}{u}+0\]

  24. UsukiDoll
    • one year ago
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    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
    • one year ago
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    well actually it is a definition in most calculus books the following: \[\ln(x)=\int\limits_1^x \frac{1}{t} dt\]

  26. anonymous
    • one year ago
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    okay i get it so now i just isolate y right?

  27. freckles
    • one year ago
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    you can if you are asked to find an explicit form for y

  28. anonymous
    • one year ago
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    it just says solve the differential equation ... so i assume i just solve for y

  29. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    okay but then how do i get rid of the absolute value ?

  32. UsukiDoll
    • one year ago
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    \[\large \ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K\] \[\large e^{ln|y|}=e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K}\] \[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K}\]

  33. freckles
    • one year ago
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    \[e^{\ln|y|}=|y|\]

  34. anonymous
    • one year ago
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    these are my answer choices by the way and none of the above

  35. freckles
    • one year ago
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    as long as y is not 0 of course

  36. freckles
    • one year ago
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    e^K=C @Jdosio

  37. anonymous
    • one year ago
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    ugh i thought your K was your way of saying Constant

  38. UsukiDoll
    • one year ago
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    forgot that part... e^K = C substitution.

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

  40. UsukiDoll
    • one year ago
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    \[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}e^{K} \] \[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}C\]

  41. freckles
    • one year ago
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    @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
    • one year ago
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    okay so wtf is K then

  43. freckles
    • one year ago
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    e^k is a constant

  44. freckles
    • one year ago
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    C is a constant

  45. freckles
    • one year ago
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    you are still replacing an unknown constant with another unknown constant

  46. anonymous
    • one year ago
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    why did you use K instead of c then

  47. freckles
    • one year ago
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    because you can

  48. freckles
    • one year ago
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    you can mean any letter to be a constant

  49. freckles
    • one year ago
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    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
    • one year ago
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    you can have any letters as a constant tbh. but it's usually C

  51. anonymous
    • one year ago
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    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
    • one year ago
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    e^K =C doesn't imply K=C

  53. UsukiDoll
    • one year ago
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    I usually just leave the solution alone...unless it begs to be changed .

  54. anonymous
    • one year ago
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    okay so then its just to keep track of which constant is which , okay that sort of makes sense

  55. freckles
    • one year ago
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    it seems funny to me to replace e^C with C

  56. freckles
    • one year ago
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    it is like replace e^5 with 5

  57. freckles
    • one year ago
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    even though e^5 doesn't equal 5

  58. anonymous
    • one year ago
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    i find it soothing to just be able to take out constants like that

  59. freckles
    • one year ago
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    but yeah you could have just said e^C then replace it with C i think some people do that

  60. freckles
    • one year ago
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    take out constants?

  61. anonymous
    • one year ago
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    combine

  62. anonymous
    • one year ago
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    anyway thank you very much for the help guys i wish i could give you both a medal :D

  63. anonymous
    • one year ago
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    ugh who do i give it too D:

  64. UsukiDoll
    • one year ago
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    I gave one to freckles already... so......?

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