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Math Help - [SOLVED] Help with understanding differential equations + graphs

  1. #1
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    [SOLVED] Help with understanding differential equations + graphs

    Hey,
    Can someone help me approach this problem correctly please:

    Find the sol. of the differential equation
    e^(-y)dy/dt+2*cos(t)=0
    My answer: y= -ln(2*sin(x)+C)
    a)sketch three soltuions corresponding to severeal values of constant of integration, C. Does every value of the consnt of intergration corespond to a solution curve? If not, which Cs do occur?
    I graphed 3 solutions, but I don't get the second part.

    b)Explain if all solutions have the same domain?
    I can answer this if i completely get part a

    c) sketch the direction feild assosiated with this equation and superimpose your sketches of the solution curves on the direction feild. (suggestion for sketching direction feild: sketch several points along the line t=0, then at the coresponding points along the lines t=pi/6, pi/3, pi/2, etc.
    i dont know how to start part c

    Ill appreciate any kind of help; thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    ok

    THe answers to the diff eq. are called integral curves...and yes each different C represents a different integral curve...

    you said you can answer two

    and for the third one start with isoclines take e^(-y)dy/dt+2*cos(t)=0
    so ln(e^(-y)dy/dt))=ln(-2cos(t)) and then -y=-ln(y')+ln(-2cos(t)) ...now call y' c...now pick a value for c..say one when c=1 y=-y=-ln(1)+ln(-2cos(t)) =ln(-2cos(t)) and you graph ln(-2cos(t)) and every point on that line will have a direction of 1...now pick more c's and repeat
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  3. #3
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    Quote Originally Posted by coe236 View Post
    Hey,
    Can someone help me approach this problem correctly please:

    Find the sol. of the differential equation
    e^(-y)dy/dt+2*cos(t)=0
    My answer: y= -ln(2*sin(x)+C)
    a)sketch three soltuions corresponding to severeal values of constant of integration, C. Does every value of the consnt of intergration corespond to a solution curve? If not, which Cs do occur?
    I graphed 3 solutions, but I don't get the second part.

    b)Explain if all solutions have the same domain?
    I can answer this if i completely get part a

    c) sketch the direction feild assosiated with this equation and superimpose your sketches of the solution curves on the direction feild. (suggestion for sketching direction feild: sketch several points along the line t=0, then at the coresponding points along the lines t=pi/6, pi/3, pi/2, etc.
    i dont know how to start part c

    Ill appreciate any kind of help; thanks
    a) e^{-y} \frac{dy}{dt} + 2 \cos t = 0 \Rightarrow e^{-y} \frac{dy}{dt} = -2 \cos t \Rightarrow \int e^{-y} \, dy = -2 \int \cos t \, dt.

    Therefore - e^{-y} = -2 \sin (t) + A \Rightarrow y = - \ln (2 \sin (t) - A).

    So your answer is correct - apart from the fact that you have x instead of t in it - (my A is equal to -C which is OK snce both represent arbitrary constants).

    Note that the solution is only defined for 2 \sin (t) + C > 0 \Rightarrow C > - 2 \sin t. So if you want the domain to be all real numbers, that is, -oo < t < +oo, then you can only have values of C > 2 since -1 \leq \sin t \leq 1.

    Note however that since -1 \leq \sin t \leq 1, there will be no value of t for which 2 \sin (t) + C > 0 if C \leq -2 ......

    So the answer is that only values C > -2 are allowed ......

    ----------------------------------------------------------------------------

    b) If C > 2 then all solutions will have the same domain. But if C \leq 2 then the domain will be restricted so that 2 \sin (t) + C > 0. What happens in the latter case will depend on the value of C. Eg. if C = 1, then you require 2 \sin (t) + 1 > 0 \Rightarrow \sin t > -\frac{1}{2} .......

    So all solutions don't have the same domain. The domain depends on the value of C.

    ----------------------------------------------------------------------------------

    c) You choose a value of t, get the value of y from the solution, and then calculate the value of dy/dt at that point. Draw a little arrow starting at the point that has the same gradient as the value of dy/dt at that point.

    Do this for different value of t. Your notes should have examples of how to do this.
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  4. #4
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    thankyou mr fantastic, you're the best
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