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Math Help - Slope Fields/Differential Equations

  1. #1
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    Slope Fields/Differential Equations

    a) Find the general solution of:

    e^(-y)*(dy/dt)+2 cos t=0.

    I'm quite sure this is y=-ln(2 sin(t)+C)

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

    Is it just that you need 2 sin(t)+C>0 and so it occurs if C>-2 sin (t)? Or is there something more I can say here? Obviously no C will exist that is less than -2 and also all Cs will exist if C is greater than 2, but between -2 and 2 it depends on the value of t?

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

    Again, I'd say no. For instance -ln(2 sin(t)+2) has domain sin(t)>-1 and -ln (2 sin (t)+3) has domain sin(t)>-3/2, but actually this is the same. I am not quite sure how to explain this one, but quite sure it is NO.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    a) Find the general solution of:

    e^(-y)*(dy/dt)+2 cos t=0.

    I'm quite sure this is y=-ln(2 sin(t)+C)

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

    Is it just that you need 2 sin(t)+C>0 and so it occurs if C>-2 sin (t)? Or is there something more I can say here? Obviously no C will exist that is less than -2 and also all Cs will exist if C is greater than 2, but between -2 and 2 it depends on the value of t?

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

    Again, I'd say no. For instance -ln(2 sin(t)+2) has domain sin(t)>-1 and -ln (2 sin (t)+3) has domain sin(t)>-3/2, but actually this is the same. I am not quite sure how to explain this one, but quite sure it is NO.
    Your solution for a) is correct.

    For b) you need to analyze three different cases

     C < -2, \quad C \in [-2,2], \quad C > 2

    As you said the first cases has not real solutions. The 2nd case will exists for some t but not all and the last cases will exists for all t.

    This will help you with part C as well.
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