11_pdfsam_math 54 differential equation solutions odd

11_pdfsam_math 54 - Exercises 1.2 C =2 10 C =1 C =0.5 C =0 0.5 0.5 C =0.5 C =1 10 C =2 Figure 1A Graphs of the functions y = Ce3x 1 for C = 0 0.5 1

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Exercises 1.2 –10 10 –0.5 0.5 C =2 C =1 C =0 . 5 C =0 C = 0 . 5 C = 1 C = 2 Figure 1–A : Graphs of the functions y = Ce 3 x +1for C =0, ± 0 . 5, ± 1, and ± 2. 17. DiFerentiating φ ( x ), we ±nd that φ 0 ( x )= ± 2 1 ce x ² 0 = ³ 2(1 ce x ) 1 ´ 0 =2 ( 1) (1 ce x ) 2 (1 ce x ) 0 ce x (1 ce x ) 2 . (1.2) On the other hand, substitution of φ ( x )for y into the right-hand side of the given equation yields φ ( x )( φ ( x ) 2) 2 = 1 2 2 1 ce x ± 2 1 ce x 2 ² = 2 1 ce x ± 1 1 ce x 1 ² = 2 1 ce x 1 (1 ce x ) 1 ce x = 2 ce x (1 ce x ) 2 , which is identical to φ 0 ( x ) found in (1.2).
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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.

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