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Math Help - initial value problem and critical value of a0.

  1. #1
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    initial value problem and critical value of a0.

    I tried extremely hard on this one. Please help me.

    Given: (sint)y' + (cost)y=e^t; y(1)=a, 0<t<pi

    a. find the initial value problem and find the critical value a0 exactly. '
    so to make this into standard form, I divided by sint.
    so: y' + (cost)y/(sint)=e^t/(sint)
    μ(t)= e^(ln(sint))
    Multiplying both sides by e^(ln(sint)): e^(ln(sint))y' + cost/sint*e^(ln(sint))y=e^t/sint*e^(ln(sint)). Then I got stuck....8(

    b. describe the behavior of the solution corresponding to the initial value a0.

    c. Draw the direction field of the given differential equation. How do the solutions behave as t goes to 0? Does the behavior depend of the choice of the initial value a? Why or why not? Let a0 be the value for a for which the transition from one type of behavior to another occurs. Estimate the value of a0.
    Last edited by mr fantastic; April 6th 2011 at 04:03 AM. Reason: Restored deleted question.
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  2. #2
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    Try using the Integrating Factor method.

    \frac{dy}{dt}+ycot(t)=\frac{e^t}{sin(t)}
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  3. #3
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    Or even quicker: notice that the LHS is the total derivative

    \dfrac{d}{dt}(\sin(t)y).

    Then integrate immediately. Essentially, this is what your integrating factor method is going to do anyway. That's what the integrating factor method is designed to do.
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  4. #4
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    i din't get the method you specified?
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  5. #5
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    Quote Originally Posted by amul28 View Post
    i din't get the method you specified?
    To whom are you replying?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    To whom are you replying?
    sorry, i am replying to you.
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  7. #7
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    Well, if you follow the program I've laid out for you, you will obtain

    \dfrac{d}{dt}(\sin(t)y)=e^{t}. To see why this is equivalent to the original DE, just take the derivative on the LHS to get

    \sin(t)\dot{y}+\cos(t)y.

    Hence, if you integrate with respect to time, you get

    \displaystyle\int\frac{d}{dt}(\sin(t)y)\,dt=\int e^{t}\,dt, or, by using the Fundamental Theorem of the Calculus, you get

    \sin(t)y=e^{t}+C. Can you continue from here? Is everything clear now?
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  8. #8
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    Can you please solve the whole thing? I really don't get it.
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  9. #9
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    Quote Originally Posted by Taurus3 View Post
    Can you please solve the whole thing? I really don't get it.
    That isn't the way this forum operates. I've essentially solved the DE for you. The business about critical points is something for which you'll need to review your notes, the definition of critical point, etc.
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