More questions on 2.4 #17

Professor: For part c of #17, I believe I have found C for y(0) = 2.
However, I am not certain how to solve for y when t = 5. I have tried putting it into Sage, but the curve it has generated doesn’t seem accurate because it does not approach 4 as it should. I am not sure what Sage is doing wrong, and I cannot finish part c or d if I cannot solve for y.
My apologies for bothering you on this question again. Any help you can provide will be much appreciated.

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One comment on “More questions on 2.4 #17

  1. We solved in class (or at least highlighted the steps) and we found the implicit solution.
    Use the initial condition to find c.
    I don’t remember the exact numbers, but say the implicit solution is something like
    1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|=-.25*t+c and suppose that you found c=0.3
    For part (c), you plug in t=5 on the right-hand side of the equation and you get:
    1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|=-.25*5+0.3.
    You have to find y.
    One way is to graph the left-hand side which is a function of y:
    1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|
    Also graph the constant function -.25*5+0.3 which appears on the right-hand side.
    Plot these two graphs together and observe where they intersect. Zoom-in until you have enough precision. That is the y-value you are looking for.

    For part (d), we said that you look what happens to the solution with initial condition y(0)=2 and the solution with initial condition y(0)=6. Every other solution is between these two.
    I will focus on the y(0)=6 case. The other one is similar (and easier).
    1. Use y(0)=6 to find c. The steps are similar to what you did for part (c).
    2. Once you have found c, plug c and y=4.05 into the equation
    1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|=-.25*t+c.
    You get
    1/12*ln|1-4.05|*(-1)+4/3*ln|1-4.05/4|*(-4)+ln|4.05|=-.25*t+c.
    Then solve for t.
    Repeat for y(0)=2.

    Does this help?

    Like

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