This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: u ± u1 2 ² = 1 This is separable: u1 = 2 u ⇒ u = 2 u + 1 = 2 + u u ⇒ Z u 2 + u du = Z dx Note that u 2 + u = u + 22 2 + u = 12 2 + u Then Z ± 12 2 + u ² du = Z dx ⇒ u2 ln  2 + u  = x + C Substitute back u = x + 2 y : ( x + 2 y )2 ln  2 + x + 2 y  = x + C Apply IC: 0 + 2(2)2 ln  2 + 0 + 2(2)  = 0 + C42 ln 2  = C The solution is given in implicit form as 2 y2 ln  2 + x + 2 y  + 4 + 2 ln 2 = 0...
View
Full Document
 Fall '06
 wirkus
 Derivative, Trigraph, UCI race classifications, Substitute good, Logarithm, Hugh Jackman

Click to edit the document details