An Operator Method for Solving Second Order Differential Equations, Part 2

In a previous post we discussed an operator method for solving certain second order ordinary differential equations. In this post I’ll explore this operator method a little further.

I first learned about this method from an old book, Higher Mathematics for Engineers and Physicists, by Ivan S. Sokolnikoff and Elizabeth S. Sokolnikoff, McGraw-Hill, 1941. I discovered the book while browsing in a used book store a few years ago, the last time I taught differential equations. You will find some of the ideas behind this post on pages 287 ff.

Consider a first order linear differential equation, with constant coefficients, which is of the form

y^{\prime} + ay = f(x)

where a is a constant. A well-known method of solution is to multiply each term by a suitable integrating factor, e^{ax} in this case, to obtain

e^{ax}y^{\prime} + ae^{ax}y = e^{ax}f(x)

Then the left side can be written as the derivative of a product:

\dfrac{{\rm d}}{{\rm d}x} \, ( e^{ax}y ) = e^{ax}f(x)

Integrate both sides, and Voila, you have the solution:

e^{ax}y = \int e^{ax}f(x) \, {\rm d} x

y = e^{-ax}\int e^{ax}f(x) \, {\rm d} x

It’s worthwhile trying out the final solution formula for a number of examples, just to confirm that it works in these cases. You can check your work by running the proposed solution through the differential equation to see if the left side really does equal the right side. Remember that a constant of integration will appear on the right side of the formula.

This brings up another standard approach to solving linear differential equations with constant coefficients, which is typically applied to second order (and higher) differential equations. One first pretends that the function f(x) is identically zero, and solves the resulting equation, called the homogeneous equation. (The term “homogeneous” is one of those irksome terms in mathematics that stands for several different concepts, depending on context.) The result is what is called “the general solution to the homogeneous equation,” and it contains a constant of integration. Then one determines one single solution (i.e., not containing any constant of integration), called “the particular solution” to the non-homogeneous equation (i.e., now including the actual function f(x) on the right side of the equation). The general solution to the original (nonhomogeneous) differential equation is the sum of the two solutions just mentioned.

Now let’s look at what the Sokolnikoffs wrote: Let’s go back and consider the equation we started off with:

y^{\prime} + ay = f(x)

Express the equation in “operator” form:

(D + a)y = f(x)

If we pretend that the operator (D + a) is mulitplied by y (which is not so; it operates on y), then we are encouraged to “solve” the differential equation by division:

y = \dfrac{1}{D + a} \; f(x)

What on earth can it mean to divide by a differential operator?? The Sokolnikoffs say that “it is convenient to associate [this] symbol” with the integral mentioned earlier. That is, the Sokolnikoffs are saying, in effect, “Look: We’ve already solved this problem above, so why don’t we just define the meaning of this strange-looking operation as the solution that we already worked out?” Seems reasonable:

\dfrac{1}{D + a} \; f(x) = e^{-ax}\int e^{ax}f(x) \, {\rm d} x

where the Sokolnikoffs emphasize that it is “convenient” to omit the constant of integration from the right side of this definition.

Why is it convenient to omit the constant of integration from the right side of the previous relation? Because this fits in with the standard method for determining the general solution for differential equations of this type. One first determines the general solution of the associated homogeneous equation, then one determines a particular solution for the nonhomogeneous equation. Their definition is convenient because they now have a nice neat formula for the particular solution.

For a second order equation of this type, one can write something like

(D + b)(D + a)y = f(x)

for which a particular solution can be written as (note the order of the factors!):

y = \dfrac{1}{D + a} \; \dfrac{1}{D + b} \; f(x)

This can be expressed in terms of integrals as

y = e^{-ax}\int [ e^{ax} \cdot e^{-bx}\int e^{bx}f(x) \, {\rm d} x ] \, {\rm d} x

which results in a similar nice, neat formula:

y = e^{-ax}\int [ e^{(a - b)x} \int e^{bx}f(x) \, {\rm d} x  ] \, {\rm d} x

For higher order linear differential equations with constant coefficients, for which the differential operator is factorable, one just strings together more integrals.

What is the practical use of the Sokolnikoff’s method? Well, after playing around with the method for a while, during the class I taught a few years ago, we concluded that it is a very useful alternative to the method of variation of parameters for determining particular solutions. However, when the method of undetermined coefficients is applicable, it might be easier just to stick with it, at least in the second-order case.

But, as always, it’s worth playing with the method yourself, both to get a feel for it, and to decide for yourself what its worth is and when it’s useful.

* * *

Click here to read An Operator Method for Solving Second Order Differential Equations, Part 3: Wild Speculation.

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About Santo D'Agostino

I have taught mathematics and physics since the mid 1980s. I have also been a textbook writer/editor since then. Currently I am working independently on a number of writing and education projects while teaching physics at my local university. I love math and physics, and love teaching and writing about them. My blog also discusses education, science, environment, etc. https://qedinsight.wordpress.com Further resources, and online tutoring, can be found at my other site http://www.qedinfinity.com
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3 Responses to An Operator Method for Solving Second Order Differential Equations, Part 2

  1. Pingback: An Operator Method for Solving Second Order Differential Equations | QED Insight

  2. Wisdom says:

    Solve examples of the form (D+a)y=cosx and also ,(D+a)y=sinx
    Then send to my email please

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