math1005problem3solr

math1005problem3solr - u 2 = Z y 1 f W dx = 1 5 Z x 4 ln( x...

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MATH 1005A - Solutions 3 Find the general solution: 1. x 2 y 00 +2 xy 0 ¡ 2 y =0 ;x> 0. This is an Euler equation. x = e t and y ( x )= z ( t ) ) z 00 + z 0 ¡ 2 z =0 :z = e rt ) r 2 + r ¡ 2=0 ) ( r +2)( r ¡ 1) = 0 ) z = c 1 e ¡ 2 t + c 2 e t ) y = c 1 x ¡ 2 + c 2 x = c 1 x 2 + c 2 x . 2. x 2 y 00 +5 xy 0 +4 y =0 ;x> 0. This is an Euler equation. x = e t and y ( x )= z ( t ) ) z 00 +4 z 0 +4 z =0 :z = e rt ) r 2 +4 r +4=0 ) ( r +2) 2 =0 ) z = c 1 e ¡ 2 t + c 2 te ¡ 2 t ) y = c 1 x 2 + c 2 ln( x ) x 2 . 3. x 2 y 00 +4 xy 0 +4 y =0 ;x> 0. This is an Euler equation. x = e t and y ( x )= z ( t ) ) z 00 +3 z 0 +4 z =0 :z = e rt ) r 2 +3 r +4=0 ) r = ¡ 3 § p 9 ¡ 16 2 = ¡ 3 2 § p 7 2 i ) z = e ¡ 3 2 t " c 1 cos à p 7 2 t ! + c 2 sin à p 7 2 t !# ) y = x ¡ 3 2 " c 1 cos à p 7 2 ln( x ) ! + c 2 sin à p 7 2 ln( x ) !# . 4. x 2 y 00 ¡ 6 y = x 3 ln( x ) ;x> 0. x 2 y 00 ¡ 6 y = 0 is an Euler equation. x = e t and y ( x )= z ( t ) ) z 00 ¡ z 0 ¡ 6 z =0 : z = e rt ) r 2 ¡ r ¡ 6=0 ) ( r ¡ 3)( r +2)=0 ) z 1 = e 3 t ;z 2 = e ¡ 2 t ) y 1 = x 3 ;y 2 = x ¡ 2 ;y h = c 1 x 3 + c 2 x ¡ 2 :f ( x )= x 3 ln( x ) x 2 = x ln( x ) ; W [ y 1 ;y 2 ]= ¯ ¯ ¯ ¯ x 3 x ¡ 2 3 x 2 ¡ 2 x ¡ 3 ¯ ¯ ¯ ¯ = ¡ 5 ;u 1 = ¡ Z y 2 f W dx = 1 5 Z ln( x ) x dx = 1 10 [ln( x )] 2 ;
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Unformatted text preview: u 2 = Z y 1 f W dx = 1 5 Z x 4 ln( x ) dx = 1 25 x 5 ln( x ) + 1 25 Z x 4 dx = 1 25 x 5 ln( x ) + 1 125 x 5 ) y p = u 1 y 1 + u 2 y 2 = 1 10 [ln( x )] 2 x 3 + 1 25 x 5 ln( x ) + 1 125 x 5 x 2 = 1 10 [ln( x )] 2 x 3 1 25 x 3 ln( x ) + 1 125 x 3 ) y = y p + y h = 1 10 [ln( x )] 2 x 3 1 25 x 3 ln( x ) + 1 125 x 3 + c 1 x 3 + c 2 x 2 ....
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This note was uploaded on 01/18/2012 for the course MATH 1005 taught by Professor Any during the Summer '07 term at Carleton CA.

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