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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)