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Math Help - y'' + by' + 4y = 0; what effect does changing b have?

  1. #1
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    y'' + by' + 4y = 0; what effect does changing b have?

    Here's the full text of the question:

    To see the effect of changing the parameter b in the initial value problem:

    y'' + by' + 4y = 0; y(0) = 1, y'(0) = 0

    Solve for b = 5, b = 4 and b = 2 and sketch the solutions.


    So basically, in short, we have the following IVPs:

    1. y'' + 5y' + 4y = 0; y(0) = 1, y'(0) = 0
    2. y'' + 4y' + 4y = 0; y(0) = 1, y'(0) = 0
    3. y'' + 2y' + 4y = 0; y(0) = 1, y'(0) = 0

    I already worked out all of them. For #1, I got y = -\frac{1}{3}e^{4x} + \frac{4}{3}e^{-x}. For #2, I wound up with y = e^{-2x} + 2xe^{-2x}. And lastly, for #3, I got y = e^{-x}cos(\sqrt{3}x) + \frac{\sqrt{3}}{3}e^{-x}sin(\sqrt{3}x). I graphed all of these functions on my calculator to look for any similarities (I swear I didn't make any mistakes in doing so; I was extremely careful with the notation), but couldn't find any.

    My question is this: what effect does changing the coefficient of the first derivative have on the solution of a second-order, linear, homogeneous ODE with constant coefficients, especially in my particular case?
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  2. #2
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    Quote Originally Posted by BlazingFire View Post
    Here's the full text of the question:

    To see the effect of changing the parameter b in the initial value problem:

    y'' + by' + 4y = 0; y(0) = 1, y'(0) = 0

    Solve for b = 5, b = 4 and b = 2 and sketch the solutions.

    So basically, in short, we have the following IVPs:

    1. y'' + 5y' + 4y = 0; y(0) = 1, y'(0) = 0
    2. y'' + 4y' + 4y = 0; y(0) = 1, y'(0) = 0
    3. y'' + 2y' + 4y = 0; y(0) = 1, y'(0) = 0

    I already worked out all of them. For #1, I got y = -\frac{1}{3}e^{4x} + \frac{4}{3}e^{-x}. For #2, I wound up with y = e^{-2x} + 2xe^{-2x}. And lastly, for #3, I got y = e^{-x}cos(\sqrt{3}x) + \frac{\sqrt{3}}{3}e^{-x}sin(\sqrt{3}x). I graphed all of these functions on my calculator to look for any similarities (I swear I didn't make any mistakes in doing so; I was extremely careful with the notation), but couldn't find any.

    My question is this: what effect does changing the coefficient of the first derivative have on the solution of a second-order, linear, homogeneous ODE with constant coefficients, especially in my particular case?
    The technique is to look for a solution of the form y = A e^{\lambda x}. When you substitute this into the DE and siplify you get the auxillary equation:

    \lambda^2 + b \lambda + 4 = 0.

    The number of solutions for \lambda and hence the type of solution to the DE will depend on the discriminant of this quadratic ....

    1. \Delta > 0.

    2. \Delta = 0.

    3. \Delta < 0.

    So think about the value of b for each of these three cases ....
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