# 285pex1 - v = y=x |{z memorize y = vx y = v xv 1 Bernoulli...

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Math. 285(D2): Exam 1 Monday, February 14, 2011 Topics: Chapter 1: 1. Sec.1.2: Integrals as general and particular solutions; di/erential equa- tions dy dx = f ( x ) : and uniqueness of solutions (Theorem 1, Sec. 1.3). Be able to apply the existence and uniqueness theorem for the ODE y 0 = f ( x;y ) and be able to handle cases when existence or uniqueness parts are not applicable. 3. Sec. 1.4: Separable equations: dy dx = f ( x ) g ( y ) ; applications: Newton±s law of cooling, Toricelli±s law, continuously compounded interest. dy dx + P ( x ) y = Q ( x ) ; integrating factor ( x ) = e R p ( x ) dx | {z } : memorize Mixture problems. 5. Sec. 1.6: Substitution method. Homogeneous equations: dy dx = F y x ± ; substitution:

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Unformatted text preview: v = y=x | {z } memorize ) y = vx ) y = v + xv : 1 Bernoulli equations: dy dx + P ( x ) y = Q ( x ) y n ;n 6 = 0 ; 1; substitution: v = y 1 & n | {z } memorize ) y = v 1 1 & n ) y = 1 1 & n v 1 1 & n & 1 v = 1 1 & n v n 1 & n v and exact equations M ( x;y ) dx + N ( x;y ) dy = 0 ; M y = N x . Di/erential forms, exact di/erential forms. Conditions for exactness for di/erential forms. Reducible second order equations: I. F ( x;y ;y 00 ) = 0 (dependent variable is missing); substitution: p ( x ) = y ( x ) ) y 00 ( x ) = p ( x ) : II. F ( y;y ;y 00 ) = 0 (independent variable is missing; substitution: p ( y ) = y ) y 00 = dp dy dy dx = p ( y ) p ( y ) : 2...
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285pex1 - v = y=x |{z memorize y = vx y = v xv 1 Bernoulli...

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