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anonymous

  • 5 years ago

how do you do this separable equations problem: dx/dt = te^t x(0)=1

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  1. anonymous
    • 5 years ago
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    \[\frac{dx}{dt}=te^t \rightarrow dx=te^tdt \rightarrow \int\limits_{}{}dx=\int\limits_{}{}te^tdt\]

  2. anonymous
    • 5 years ago
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    \[\frac{dx}{dt}=te^t \rightarrow dx=te^tdt \rightarrow \int\limits_{}{}dx=\int\limits_{}{}te^tdt\]

  3. anonymous
    • 5 years ago
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    This is just\[x=e^t(t-1)+c\]

  4. anonymous
    • 5 years ago
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    To find c, you have\[x(0)=e^0(0-1)+c=1 \rightarrow c=2\]

  5. anonymous
    • 5 years ago
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    So your equation is\[x(t)=e^t(t-1)+2\]

  6. anonymous
    • 5 years ago
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    how did you get e^t(t-1) is it necessary to use the product rule for that integration or what?

  7. anonymous
    • 5 years ago
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    You will use integration by parts.

  8. anonymous
    • 5 years ago
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    could you show me how to do that?

  9. anonymous
    • 5 years ago
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    Set u=t and dv=e^tdt

  10. anonymous
    • 5 years ago
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    ok

  11. anonymous
    • 5 years ago
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    I'll write it and scan - it will be quicker.

  12. anonymous
    • 5 years ago
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    alright thanks i'm trying it now

  13. anonymous
    • 5 years ago
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    i'm stuck

  14. anonymous
    • 5 years ago
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    I'm scanning

  15. anonymous
    • 5 years ago
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    alright thanks

  16. anonymous
    • 5 years ago
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    Just have to wait a min, since the software for my scanner is a pain

  17. anonymous
    • 5 years ago
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    alright cool

  18. anonymous
    • 5 years ago
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    do you think you could help me out with another one after this?

  19. anonymous
    • 5 years ago
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    Maybe - I'm supposed to be getting ready for uni.

  20. anonymous
    • 5 years ago
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    alright well its dy/dx = (xy)^(1/2) and i just wasn't sure how you set that one up

  21. anonymous
    • 5 years ago
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    Just split the factors up using power laws.

  22. anonymous
    • 5 years ago
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    so you would end up with dy/sqrt(y) = sqrt x dx

  23. anonymous
    • 5 years ago
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    \[\frac{dy}{dx}=(xy)^{1/2} = x^{1/2}y^{1/2} \rightarrow \frac{dy}{y^{1/2}}=x^{1/2}dx\]

  24. anonymous
    • 5 years ago
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    then after you integrate would you have 2 sqrt y = 2/3 x^(3/2)

  25. anonymous
    • 5 years ago
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  26. anonymous
    • 5 years ago
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    Here's the scan for the first one.

  27. anonymous
    • 5 years ago
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    and when it's all said and done i had \[y= (x ^{3/2}/3)^2 + (c/2)^2\]

  28. anonymous
    • 5 years ago
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    i feel like thats not right at all

  29. anonymous
    • 5 years ago
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    \[\int\limits{}{}\frac{dy}{y^{1/2}}=\frac{y^{1/2}}{1/2}=2y^{1/2}\]

  30. anonymous
    • 5 years ago
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    and

  31. anonymous
    • 5 years ago
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    \[\int\limits_{}{}x^{1/2}dx = \frac{x^{3/2}}{3/2}=\frac{2}{3}x^{3/2}+c\]

  32. anonymous
    • 5 years ago
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    I added the constant then.

  33. anonymous
    • 5 years ago
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    Equate the two.

  34. anonymous
    • 5 years ago
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    \[2y^{1/2}=\frac{2}{3}x^{3/2}+c\]

  35. anonymous
    • 5 years ago
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    Divide by 2 to clean up:\[y^{1/2}=\frac{1}{3}x^{3/2}+c\]

  36. anonymous
    • 5 years ago
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    You don't have to write c/2 since c is just a constant - constants just keep absorbing constants since in the end, their value is determined by initial conditions and you'll end up with the same number.

  37. anonymous
    • 5 years ago
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    ohhhh ok i think i got it since c is a constant i dont need to divide it

  38. anonymous
    • 5 years ago
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    haha ya thanks alot

  39. anonymous
    • 5 years ago
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    np

  40. anonymous
    • 5 years ago
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    i really appreciate it

  41. anonymous
    • 5 years ago
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    become a fan then :)

  42. anonymous
    • 5 years ago
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    how?

  43. anonymous
    • 5 years ago
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    there should be a link next to my name in this area we're typing in. It's a blue link that says, "Become a fan"

  44. anonymous
    • 5 years ago
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    Should have a 'thumbs up' icon...

  45. anonymous
    • 5 years ago
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    ya i dont see it?

  46. anonymous
    • 5 years ago
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    it just has hero and the 110 fans

  47. anonymous
    • 5 years ago
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    refresh the page...sometimes that helps

  48. anonymous
    • 5 years ago
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    it just has hero and the 110 fans

  49. anonymous
    • 5 years ago
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    yeah, the site's awkward

  50. anonymous
    • 5 years ago
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    ah

  51. anonymous
    • 5 years ago
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    sweet i got it

  52. anonymous
    • 5 years ago
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    happy travels )

  53. anonymous
    • 5 years ago
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    thanks man i appreciate it

  54. anonymous
    • 5 years ago
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    later

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