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Unformatted text preview: y = vx 1 dy dx = v + x dv dx v + x dv dx = g ( v ) Seperating the variables dv vg ( v ) + dx x = 0 Z dv vg ( v ) + Z dx x = c Let F ( v ) = Z dv vg ( v ) Then, the solution takes the form F ( y x ) + ln  x  = c Example 1 Solve the equation ( x 23 y 2 ) dx + (2 xy ) dy = 0 dy dx = 3 y 2x 2 2 xy 3 y 2x 2 2 xy = 3 y 2 xx 2 y = 3 2 ( y x )1 2 ( 1 y x ) dy dx = 3 2 ( y x )1 2 ( 1 y x ) The right part is of the form g ( y x ) . So, the diﬀerential equation is homogeneous. We can now proceed to solve the equation. ( x 23 y 2 ) dx + (2 xy ) dy = 0 dy dx = 3 y 2 xx 2 y 2 Let y = vx v + x dv dx =1 2 v + 3 v 2 x dv dx =1 2 v + v 2 x dv dx = v 21 2 v Seperate the variables 2 v v 21 dv = dx x ln ± ± v 21 ± ± = ln  x  + ln  c  ± ± v 21 ± ± =  cx  Replace v by y x ± ± ± ± y 2 x 21 ± ± ± ± =  cx  y 2x 2 = cx 3 3...
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This note was uploaded on 08/03/2011 for the course ECON 101 taught by Professor Profeessor during the Spring '11 term at Aachen University of Applied Sciences.
 Spring '11
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