# 4-9.Cauchy - Cauchy-Euler equation Definition A linear...

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Cauchy-Euler equation Definition . A linear differential equation of the form a n x n y ( n ) + a n 1 x n 1 y ( n 1 ) + + a 1 x y ' + a 0 y = g ( x ) where the coefficients a n ,a n 1 ,…,a 0 are constants, is called a Cauchy-Euler equation. We start the discussion with a detailed examination of the homogeneous second- order a x 2 y ' ' + bx y ' + cy = 0 ( 2 ) Also, we can solve the non-homogeneous equation a x 2 y ' ' + bx y ' + cy = g ( x ) Method of solution. We try a solution of the form y = x m , where m is to be determined. When we substitute y = x m , to the second-order equation becomes am ( m 1 ) x m + bm x m + c x m = 0 m 2 am + bm + c a ¿ x ¿ ¿ am 2 + ( b a ) m + c = 0 ( 3 ) The last equation is called the characteristic equation of the differential equation (2).

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Case I: Let m 1 and m 2 denote real and distinct roots of (3). Then general solution of (2) is y = C 1 x m 1 + C 2 x m 2 Example 1. Solve x 2 y ' ' 2 x y ' 4 y = 0 Solution. a = 1, b =− 2, c =− 4 m 2 3 m 4 = 0 D = 25 m 1 = 3 5 2 =− 1 and m 2 = 3 + 5 2 = 4 So y = C 1 x 1 + C 2 x 4 Case II: Let m 1 and m 2 be real roots and m 2 = m 1 . The general solution of (2) is y = C 1 x m 1 + C 2 x m 1 lnx Example 2. Solve 4 x 2 y' ' + 8 x y ' + y = 0 Solution. a = 4, b = 8, c = 1 4 m 2 + 4 m + 1 = 0 then m 2 = m 1 = 1 2 So y = C 1 x 1 2 + C 2 x 1 2 lnx Case III: Conjugate complex roots.
If the roots (3) m 1 = α + βi and m 2 = α βi , where α , β > 0 . then general solution is y = x α ( C 1 cos ( βlnx ) + C 2 sin ( βlnx ) ) Example 3. Solve 4 x 2 y' ' + 17 y = 0 Solution. a = 4, b = 0, c = 17 4 m 2 4 m + 17 = 0 D = 16 4 4 17 = 16 ( 1 17 ) =− 256 m 1 = 4 + 16 i 8 = 1 2 + 2 i and m 2 = 1 2 2 i ,α = 1 2 = 2 So y = x 1 2 ( C 1 cos ( 2 lnx ) + C 2 sin ( 2 lnx ) ) Example 4. Solve x 2 y ' ' 3 x y ' + 3 y = 2 x 4 e x Solution. a = 1, b =− 3, c = 3 m 2 4 m + 3 = 0 D = 16 12 = 4 m 1 = 1 and m 2 = 3 , So y c = C 1 x + C 2 x 3 . Dividing by x 2 given equation we have y' ' 3 x y ' + 3 x 2 y = 2 x 2 e x With the identifications y 1 = x , y 2 = x 3 W ( x, x ) = | x x 3 1 3 x 2 | = 2 x 3 .

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We identify f ( x )= 2 x 2 e x . Then we obtain W 1 = | 0 x 3 2 x 2 e x 3 x 2 | =− 2 x 5 e x , W 2 = | x 0 1 2 x 2 e x | = 2 x 3 e x and u 1 ' = 2 x 5 e x 2 x 3 =− x 2 e x ,u 2 ' = 2 x 3 e x 2 x 3 = e x .
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