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Transcript
```EULER EQUATIONS
Differential equations of the form:
a n x n y ( n ) + a n −1 x n −1 y ( n −1) + ...a 2 x 2 y ′′ + a1 xy ′ + a 0 y = f ( x ) (1)
represent a special case of Euler’s equation.
Equations of this sort can always be transformed into a linear differential equation with
constant coefficients by making the substitution:
x = e z or z = ln x (2)
Let’s consider the specific differential equation:
x 2 y ′′ + 2 xy ′ − l (l + 1) y = 0
(3)
When we make the substitution shown in (2) we get:
dy dy dz
dy
(4)
=
⋅
= e −z
dx dz dx
dz
the last step occurring because dx/dz = ez, so dz/dx = e-z.
The second derivative becomes:
d 2 y d ⎛ dy ⎞ d ⎛ − z dy ⎞ ⎡ d ⎛ − z dy ⎞⎤ dz ⎡ − z ⎛ dy ⎞ − z ⎛ d 2 y ⎞⎤ − z
= ⎜ ⎟ = ⎜e
= ⎢− e ⎜ ⎟ + e ⎜⎜ 2 ⎟⎟⎥ e =
⎟=
⎜e
⎟
dz ⎠ ⎢⎣ dz ⎝
dz ⎠⎥⎦ dx ⎣
dx 2 dx ⎝ dx ⎠ dx ⎝
⎝ dz ⎠
⎝ dz ⎠⎦
(5)
⎛d y⎞
⎛ dy ⎞
e − 2 z ⎜⎜ 2 ⎟⎟ − e − 2 z ⎜ ⎟
⎝ dz ⎠
⎝ dz ⎠
2
Remembering that x2 = e2z and x = ez, we can use equations (4) and (5) to rewrite (3) as:
d 2 y dy
+
− l (l + 1) y = 0
dz 2 dz
(6)
which is a simple second order homogeneous constant coefficient ordinary differential
equation.
The characteristic equation for (6) can be written as:
r 2 + r − l (l + 1) = 0 (7)
Which can be factored as:
( r − l )( r + l + 1) = 0 (8)
This gives a solution to (6) of:
y = c1e lz + c 2 e − ( l +1) z
(9)
Since x = ez, eq. (9) becomes:
y = c1 x l + c 2 x − ( l +1)
(10)
```
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