94_pdfsam_math 54 differential equation solutions odd

# 94_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 45. To obtain (17), we divide given equations: dy dx = 4 x + y 2 x y = 4+( y/x ) ( ) 2 . Therefore, the equation is homogeneous, and the substitution u = yields u + x du dx = 4+ u u 2 x du dx = u u 2 u = u 2 +3 u +4 u 2 u 2 u 2 3 u 4 du = 1 x dx Z u 2 u 2 3 u 4 du = Z 1 x dx . Using partial fractions, we get u 2 u 2 3 u 4 = 2 5 1 u 4 + 3 5 1 u +1 , and so 2 5 ln | u 4 | + 3 5 ln | u | = ln | x | + C 1 ( u 4) 2 ( u +1) 3 x 5 = C ± y x 4 ² 2 ± y x ² 3 x 5 = C ( y 4 x ) 2 ( y + x ) 3 = C. REVIEW PROBLEMS: page 81 1. Separation variables yields y 1 e y dy = e x dx ( y 1) e y dy = e x dx Z ( y 1) e y dy = Z e x dx ⇒− ( y 1) e y + Z e y dy = e x + C ( y
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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