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Unformatted text preview: AMATH 351 Homework 4 Due July 22, 2009 Section3.4 9,20,40 Section3.5 12,20,26 Section3.6 1,14 Section3.7 4,12 Section 3.4 9 y 00 + 2 y 8 y = Characteristic equation is λ 2 + 2 λ 8 = λ = 2 or 4 So the general solution is y ( t ) = C 1 e 2 t + C 2 e 4 t 20 y 00 + y = 0 , y ( π 3 ) = 2 , y ( π 3 ) = 4 Characteristic equation is λ 2 + 1 = λ = ± i So general solution is y = C 1 cos t + C 2 sin t 1 Apply the given conditions: ( C 1 2 + √ 3 2 C 2 = 2 √ 3 2 C 1 + C 2 2 = 4 = ⇒ ( C 1 = 1 + 2 √ 3 C 2 = 2 + √ 3 So y ( t ) = 1 + 2 √ 3 cos t + 2 + √ 3 sin t 40 t 2 y 00 + 4 ty + 2 y = 0 Let x = ln t , then dx dt = 1 t , dy dt = dy dx dx dt = 1 t dy dx d 2 y dt 2 = d dt 1 t dy dx = 1 t 2 dy dx + 1 t d 2 y dx 2 dx dt = 1 t 2 dy dx + 1 t 2 d 2 y dx 2 So the di erential equation can be written as t 2 1 t 2 dy dx + 1 t 2 d 2 y dx 2 + 4 t 1 t dy dx + 2 y = d 2 y dx 2 + 3 dy dx + 2 y = 0 which is a constant coe . 2nd order linear DE. So the characteristic equation is λ 2 + 3 λ + 2 = = ⇒ λ = 1 or 2 So the general solution is y ( x ) = C 1 e x + C 2 e 2 x Change back to t , we get y ( t ) = C 1...
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This note was uploaded on 01/10/2010 for the course MATH 124 taught by Professor Walker during the Spring '08 term at University of Washington.
 Spring '08
 WAlker
 Math, Calculus

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