This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **u ± u-1 2 ² = 1 This is separable: u-1 = 2 u ⇒ u = 2 u + 1 = 2 + u u ⇒ Z u 2 + u du = Z dx Note that u 2 + u = u + 2-2 2 + u = 1-2 2 + u Then Z ± 1-2 2 + u ² du = Z dx ⇒ u-2 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 + C-4-2 ln |-2 | = C The solution is given in implicit form as 2 y-2 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