Math 240 - Differential Equations

Problematic Topics

  • Orthogonal Trajectories
  • Wronskian

    Sorry, HTML is a horrible language, so formulas and spacing are not going to come out right...

    Orthogonal Trajectories

    Well, this topic is not in your textbook, but it's in the complimentary book and, besides, you are supposed to know whatever was covered in the lectures.

    Assume that we have a one-parameter family of curves, i.e., a family of curves given by an equation F(x, y, a) = 0 with one parameter a. An  orthogonal trajectory  of the family is a curve which at each point is orthogonal to the curve of the family passing through this point. Typically, orthogonal trajectories also form a one-parameter family.

    Orthogonal trajectories are found in two steps (see, e.g., Problem 3 in 2000 Midterm I).

    First, the original representation F(x, y, a) = 0 should be converted to a  differential equation   (free of parameters!)  whose integral curves are the curves of the given family. To do so, differentiate F(x, y, a) = 0 to get dF/dx+ y'dF/dy = 0 and  eliminate  a from the system

    F(x, y, a) = 0,
    dF/dx+ y'dF/dy = 0.
    (The latter is to be regarded as an algebraic system of two equations in four unknowns x, y, y', a. There is a good chance that one of the unknowns, namely, a, can be eliminated, so one can produce  one  equation G(x, y, y') = 0 in the remaining three unknowns. As you can see, this is a differential equation in x, y.)

    Now in the equation G(x, y, y') = 0 replace y with Y and y', with -1/Y', and solve the resulting equation G(x, Y, -1/Y') = 0. Its integral curves are the orthogonal trajectories of the original family.

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    The Wronskian W(x) of an n-tuple y1, ..., yn of solutions of an n-th order homogeneous linear differential equation

    y^(n) + p(x)y^(n-1) + ... + q(x)y = 0
    satisfies the equation
    W' + p(x)W = 0.
    Hence, it is given by
    W(x) = C exp(-int p(x) dx),
    and in order to find W(x) one does  not  need to know the solutions; it suffices to know their values and the values of their first (n-1) derivatives at a certain point x0 (i.e., to know that the solutions satisfy certain initial conditions). Then, substituting x = x0, one can find the constant C and, hence, W(x). See, e.g., Problem 2 in
    2000 Midterm II.

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