Quod Erat Demonstrandum

2009/10/19

Solve DE by method of substitution

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 5:35 下午

In solving ordinary differential equation

A(ax + b)^2\frac{d^2y}{dx^2} + B(ax + b)\frac{dy}{dx} + Cy = f(x) ………. (*)

(where A, B, C are constants)

we use the method of substitution, let

ax + b = e^z

Then we have \frac{dz}{dx} = ae^{-z} and hence

\frac{dy}{dx} = ae^{-z}\frac{dy}{dz}
\frac{dy}{dx} = ae^{-2z}(\frac{d^2y}{dz^2} - \frac{dy}{dz})

Therefore, (*) can be further reduced as a second order linear differential equation with constant coefficients:

aA\frac{d^2y}{dz^2} + a(B - A)\frac{dy}{dz} + Cy = g(z), where g(z) = f(\frac{e^z - b}{a})

The first question come out from students’ minds: how can you think about the substitution e^z = ax + b? Sorry, I cannot answer. This should be a great idea from someone(s) in the past, but how could he/she think about that? Well, you may explore the historical facts and tell me later. Second question, as asked by a student today morning, Chan, is it always possible to substitute e^z = ax + b? Good question. e^z is different from ax + b in general, at least e^z must be positive but ax + b may be negative.

OK, let’s consider a concrete example from a textbook.

Solve

(x + 2)^2\frac{d^2y}{dx^2} + (x + 2)\frac{dy}{dx} + y = 3x + 4 ………. (#)

In the solution, it wrote x + 2 = e^z immediately and, after a series of mechanical procedure, it yields

y = C_1(x + 2) + C_2(x + 2)\ln(x + 2) + \frac{3}{2}(x + 2)\ln^2(x + 2) - 2

Obviously, the above is valid only when x + 2 > 0.

But, there should not be any restriction on x in the equation (#).

So, what is the solution to (#) indeed? Is it simply add absolute signs to the above and yield

y = C_1(x + 2) + C_2(x + 2)\ln|x + 2| + \frac{3}{2}(x + 2)\ln^2|x + 2| - 2 ?

Urm…students, you may try on your own:

when x + 2 < 0, let x + 2 = -e^z. See what you will obtain?

OK? Do you obtain something like

y = D_1(x + 2) + D_2(x + 2)^{-1} - \frac{3}{2}(x + 2)\ln(-x-2) - 2 for x + 2 < 0 ?

(Please help me to debug, because I just did it in a hurry…)

For better setting of the question, we may post the question as

Solve

(x + 2)^2\frac{d^2y}{dx^2} + (x + 2)\frac{dy}{dx} + y = 3x + 4 for x + 2 > 0.

………………………….
[OT]

To 7B students, please refer to the following post for the question I’d mentioned in the lesson today:
https://johnmayhk.wordpress.com/2007/10/05/alpm-past-paper-1998-paper-ii-q11/

3 則迴響 »

  1. The substituition using exponential function is very useful is solving linear ODE.
    Another issue which use a similar method is the Cauchy-Euler equation.

    For further reading,
    http://en.wikipedia.org/wiki/Linear_differential_equation
    http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

    迴響 由 Justin — 2009/10/20 @ 1:34 上午 | 回覆

    • Thank you Justin for your links. As mentioned in the second webpage you’d given, the trial solution is y = x^m in solving the Euler–Cauchy equation, however, the solution turns up something like y = c_1x^m\ln(x) + c_2x^m, it may be quite puzzling.

      迴響 由 johnmayhk — 2009/10/20 @ 12:47 下午 | 回覆

  2. 我記得學過

    Liberal Studies, the Liberal Us!
    http://wp.me/PyvmP-3b

    迴響 由 lslu — 2009/10/20 @ 9:53 上午 | 回覆


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