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Math Help - Ivp!!!???

  1. #1
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    Angry Ivp!!!???

    Question:

    "y1 (t) = 1/√(1-2t) and y2 (t) = 1/√(4-2t) are solutions of dy/dt = y^3. Use the plot of the solutions to answer the following parts of this question: (a) consider the IVP consisting of the differential equation given together with the initial condition y(0)=1. Find the interval of t for which the solution to the IVP exists. (b) What can you say about the behaviour of a solution to this differential equation which has an initial condition satisfying 0.5<y(0)<1.0?"

    I've plotted these solutions and uploaded a picture of the graph. Help will be appreciated...thank you
    Attached Thumbnails Attached Thumbnails Ivp!!!???-plot.jpg  
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  2. #2
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    Quote Originally Posted by Ghostrider View Post
    Question:

    "y1 (t) = 1/√(1-2t) and y2 (t) = 1/√(4-2t) are solutions of dy/dt = y^3. Use the plot of the solutions to answer the following parts of this question: (a) consider the IVP consisting of the differential equation given together with the initial condition y(0)=1. Find the interval of t for which the solution to the IVP exists. (b) What can you say about the behaviour of a solution to this differential equation which has an initial condition satisfying 0.5<y(0)<1.0?"

    I've plotted these solutions and uploaded a picture of the graph. Help will be appreciated...thank you
    (a) y1 (t) = 1/√(1-2t) is the solution corresponding to y(0) = 1. Solutions exists for 1 - 2t > 0 => t < 1/2. You can also see this from the graph.

    (b) y2 (t) = 1/√(4-2t) is the solution corresponding to y(0) = 0.5.

    I'd discuss domain, shape and behaviour as t --> -oo.

    Note that the general solution is y = \pm \frac{1}{\sqrt{A - 2t}} and y(0) = \pm \frac{1}{\sqrt{A}}. Domain: t < \frac{A}{2}.

    \frac{1}{2} < \frac{1}{\sqrt{A}} < 1 \Rightarrow 1 < A < 4.
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