# 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 $x + 2 < 0$ ?

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.

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

• 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 | 回應

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 | 回應