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Math Help - Solutions curves

  1. #1
    Senior Member
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    Solutions curves

    Consider:
    e^(-y)(dy/dt)+2 cost t=0

    a) Solve it
    I did this and got -e^(-y)=-2sint +C, or y=-ln(2sin t)+C

    b) Does every value of C correspond to a solution curve? If not, which do and which do not and why?

    I'm not sure about this....I don't think so, since we have a negative ln....this one doesn't make too much sense to me though since 2 sin t will always be between -2 and 2, hence the only values -ln(2 sin t) can take are between -ln(-2) (which is not real) and -ln(2)...or I guess more appropriately -ln(1) and -ln(2)....

    c) Do all the solutions have the same domain? Explain.

    Probably not, but not sure.
    Thanks!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by zhupolongjoe View Post
    Consider:
    e^(-y)(dy/dt)+2 cost t=0

    a) Solve it
    I did this and got -e^(-y)=-2sint +C, or  \text{y=-ln(2sin t}  \color{red}\text{)+C}

    b) Does every value of C correspond to a solution curve? If not, which do and which do not and why?

    I'm not sure about this....I don't think so, since we have a negative ln....this one doesn't make too much sense to me though since 2 sin t will always be between -2 and 2, hence the only values -ln(2 sin t) can take are between -ln(-2) (which is not real) and -ln(2)...or I guess more appropriately -ln(1) and -ln(2)....

    c) Do all the solutions have the same domain? Explain.

    Probably not, but not sure.
    Thanks!
     e^{-y}\frac{dy}{dt} = -2\cos(t) \implies \int e^{-y}\frac{dy}{dt} dt = -\int \cos(t)dt\implies -e^{-y} = -2\sin(t)+C by the reverse chain rule.
    Therefore  y=-\ln(2\sin(t)+C) . (Note how  C is inside the logarithm.)


    For  y to exist,  -2\sin(t)+C>0 for all  t , otherwise the logarithm isn't defined.
    Thus  C > 2 yields a solution to exist for all real numbers since  -2\sin(t)+C>0 for all  t .
     -2 < C \leq 2 yields a solution for some but not all real numbers since  -2\sin(t)+C>0 for some  t .
     C \leq -2 yields no solution since  -2\sin(t)+C>0 for no  t .

    Note, I'm assuming we're strictly working in the real numbers here.
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  3. #3
    Super Member
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    Hi. I get y(t)=-\log(2\sin(t)+c) and try and keep in mind that all differential equations are equally valid in the complex plane wherever the solution is analytic (differentiable in a region). So that solution above could just as well have been written with all complex variables like:

    w(z)=-\log(2\sin(z)+z_0), everything complex, and it would still satisfy the DE in some region of analyticity where the argument to the log function is not zero so c can't be 0 when t=0. When we solve for a "real" answer, we're just taking part of the "complex" solution that gives a "real" answer but we could just as well let c=-10 and would obtain the solution:

    y(t)=-\log(2\sin(t)-10) which gives y(0)=-(\ln(10)-\pi i) and that is just as valid a solution as the real solution (in some region for which it is analytic).
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