109_pdfsam_math 54 differential equation solutions odd

# 109_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.2 An integrating factor for this equation has the form µ ( t )=exp ±Z 4 100 t dt ² = e 4 ln(100 t ) = (100 t ) 4 . Multiplying the linear equation by the integrating factor yields (100 t ) 4 dx dt +4 x (100 t ) 5 =(1 . 2)(100 t ) 4 D t ³ (100 t ) 4 x ´ =(1 . 2)(100 t ) 4 (100 t ) 4 x =1 . 2 Z (100 t ) 4 dt = 1 . 2 3 (100 t ) 3 + C x ( t )=(0 . 4)(100 t )+ C (100 t ) 4 . To Fnd the value of C , we use the initial condition x (0) = 1. Therefore, x (0) = (0 . 4)(100) + C (100) 4 =1 C = 39 100 4 = 3 . 9 × 10 7 . This means that at time t there is x ( t )=(0 . 4)(100 t ) (3 . 9 × 10 7 )(100 t ) 4 liters of nitric acid in the tank. When the percentage of nitric acid in the tank is 10%, the
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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 Berkeley.

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