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Unformatted text preview: Review Problems
⇒ ⇒ ⇒ ⇒ z 1 1 ln z 2 + 2 + √ arctan √ = − ln u + C1 2 2 2 √ z =C ln z 2 + 2 u2 + 2 arctan √ 2 √ v √ ln v 2 + 2u2 + 2 arctan =C u2 √ y−2 √ = C. ln (y − 2)2 + 2(x − 1)2 + 2 arctan (x − 1) 2 √ The initial condition, y (0) = 2, gives C = ln 2, and so the answer is ln (y − 2)2 + 2(x − 1)2 + 39. Multiplying the equation by y , we get y We substitute u = y 2 and obtain 1 1 du 2 − u= 2 dx x x which is linear and has an integrating factor µ(x) = exp Hence, d (x−4 u) = 2x−5 dx ⇒ ⇒ ⇒ Substitution y (1) = 3 yields 1 32 = − + C (1)4 2 or C= 19 . 2 101 x−4 u = x−4 y 2 = − 2x−5 dx = − x−4 +C 2 − 4 dx = x−4 . x ⇒ du 4 2 − u= , dx x x 2 1 dy − y2 = . dx x x 2 arctan y−2 √ = ln 2 . (x − 1) 2 x−4 +C 2 1 y 2 = − + Cx4 . 2 ...
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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