5.    Reduction of Order Strategies for 2nd and Higher Order ODEs

The methods presented in this section involve reducing an n'th order ODE into n first order ODEs.

5.1.    Terms Missing in ODE

When written in standard form, you may not that one or more of the coefficients multiplying the derivative terms are zero - i.e. some terms in the dependent variable are missing.  In this case, it may be usefull to introduce a new dependent variable, which equals one of the lower-order derivatives (may or may not be missing).  For example, if the ODE is second order and linear, then the solution strategy may follow the steps:
  •  substitute v=dy/dx,
  •  solve for v, but v=dy/dx
  •  solve ODE in dy/dx for y

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    NOTE:  If the ODE has the term in y(x) (i.e. y' term is missing), then an extra step is required.  The above substitution will produce an ODE in x, y and v:  dv/dx + y = 0, for example.  This is handled by noting that dv/dx equals dv/dy * dy/dx, an since we know what dy/dx is (i.e. it is v), we end up with a solvable ODE like: v*dv/dy + y = 0.

    Click here for the PDF file that shows a few examples.

    5.2.    D'Alembert's Method

    D'Alembert's Method is the first of a few ODE solution strategies that uses an initial, usually simple, fundamental solution to the ODE to find another more complex solution.  This, in turn, can be used to determine the general/complete solution to the ODE.

    The underlying principle is that, if y1(x) is a solution to an ODE, then, clearly, y2(x)=C*y1(x) is also a solution.  D'Alembert's method extends this concept to include a 'variable constant', v(x).  Thus, it is presumed that a new solution can be found such that:
        y2(x)=v(x)*y1(x)

    In order to find the more complete solution (y2), we need to determine the function v(x).  After the form of the initial 'seed' solution, y1,  is determined (or given), the function v(x) and the more complete solution is found using the following steps:
    1.    substitute the proposed solution, y2=v*y1, and all of the required derivatives into the ODE
    2.    simplify the ODE (which is now in terms of v)
    3.    solve for v(x) using 'missing term' approach
    4.    assemble y2(x)

    Click here for an example of D'Alembert's method.

    At first glance, one might assume that this could go on forever.... Now, given y2=v*y1, find y3=vNew*y2, etc.  At some point, it is reasonable to expect that this process doesn't add any more 'new' information to the solution of the ODE - That is to say, that the 'seed' solution is already the complete or general solution.

    The example above shows that this is indeed the case... we will end up with the y3 being the same as y2.

    The idea of Superposition of Solutions and Linear Independence gives us a definative way to know when to stop trying to find more solutions.
     

    5.3.    Superposition of Solutions and Linear Independence

    Two ideas come to light here:

    1)    There may be fundamental solutions to an ODE, which are building blocks to the general solution of the ODE.  That is to say, these fundamental solutions, individually, ARE NOT the general solution - only part of it

    If the ODE is linear, then the general solution is the linear combination of the fundamental set of solutions
    This simply follows from the idea of superposition: For a linear ODE, a new solution may be produced using a linear combination of two existing solutions.

    2)    A group of fundamental solutions define the fundamental set of solutions, and can be linearly combined to form the general solution to the ODE, if and only if they are linearly independent.

    For an n'th order ODE, there are n fundamental solutions in the fundamental set.  Usually, one can simply tell by inspection that two fundamental solutions are linearly indpendent... they simply look different.  Here are some examples:
     
    y1=exp(t) and y2=exp(-t)  These are linearly indpendent - they are different functions, and form a fundamental set
    y1=exp(2z) and y2=sin(z) Also linearly independent
    y1=sin(t) and y2=cos(t-pi/2) NOT linearly independent

    There is, perhaps fortunately, a more rigorous test for linear indpendence of two functions which follows from algebra.  The 'Wronskian Determinant' can be evaluated to ascertain whether or not two functions are:

  • unconditionally linearly indpendent,
  • conditionally linearly independent (giving restrictions on where they will be LI),
  • or unconditionally linearly dependent.

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    If they are dependent, the Wronskian determinant will equal zero, then their linear combination will not give the general solution.

    The Wronskian determinant is found as the determinant of y1, y2 (in the first row), and y1', y2' (in the second row). For the second order ODEs we will consider, the Wronskian is:
        W = y1*y2' - y2*y1'

    We note, in passing, that fundamental solutions to nth order ODEs can also be checked.  The Wronskian determinant would be formed using the n solutions, and up to their (n-1) derivatives.

    Click here for some examples.