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  1. #1
    la vache fantôme phantomcow2's Avatar
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    Is this integral correct?

    I'm trying to solve a differential equation but am stuck on a specific step that involves integration. Could someone please explain how the following is true?



    If we integrate both sides both sides with respect to t, I would have expected the following result:

    1/(1+t^2) - 1/(1+1^2)*y = 3*arctan(t) + C

    And factoring out the 1/(1+t^2) from the left, we would get [1/(1+t^2)](y-1). Solving for y becomes

    -3*arctan(t)*(1+t^2) +1 + C = y


    Why am I incorrect?
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  2. #2
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    I'm not 100% sure your wxy Or, xyz statement is true, but I'm not much of a math guy as far as engineers are concerned.

  3. #3
    Tiocfáidh ár Lá jfmckenna's Avatar
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    This part of your equation makes no sense: 1/(1+1^2). One would simply write that as 1/2.
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    That second 1 is a lower case T.

  5. #5
    Free @coasting RUOkie's Avatar
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    I'm pretty sure the answer is 42



    or possibly poo
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  6. #6
    Tiocfáidh ár Lá jfmckenna's Avatar
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    Quote Originally Posted by jccaclimber View Post
    That second 1 is a lower case T.
    Looks like a one to me but yeah must have been a typo.
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  7. #7
    la vache fantôme phantomcow2's Avatar
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    oops, good catch. The one should have been a lower case t. .

    RUOkie, 42 couldn't possibly be a valid solution since the solution is a general one -- it doesn't pertain to a specific number.
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  8. #8
    Portland Fred banerjek's Avatar
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    Quote Originally Posted by phantomcow2 View Post
    RUOkie, 42 couldn't possibly be a valid solution since the solution is a general one -- it doesn't pertain to a specific number.
    How can you get more general than the answer to life, the universe, and everything?

  9. #9
    You Know!? For Kids! jsharr's Avatar
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    Quote Originally Posted by phantomcow2 View Post
    oops, good catch. The one should have been a lower case t. .

    RUOkie, 42 couldn't possibly be a valid solution since the solution is a general one -- it doesn't pertain to a specific number.
    then it has to be "poo" by process of elimination. next question please.
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  10. #10
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    Are you asking how the last line in the image is correct? The integral of d/dt (X) with respect to t = X (integral of a derivative of X = X). That is y/1+t^2 in your example. You already agree with the right side integral solution so now you just need to multiply both sides by 1+t^2 to solve for y.

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