60_pdfsam_math 54 differential equation solutions odd

60_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 2 ⇒ ⇒ ⇒ d (510 − t)−6 A = 1 · (510 − t)−6 = (510 − t)−6 dt 1 (510 − t)−6 A = (510 − t)−6 dt = (510 − t)−5 + C 5 1 A = (510 − t) + C (510 − t)6 . 5 Using the initial condition, A(10) = 14.04, we compute C . 14.04 = A(10) = Therefore, A(t) = 1 85.96 1 (510 − t) − (510 − t)6 = (510 − t) − 85.96 6 5 (500) 5 510 − t 500 6 1 (510 − 10) + C (510 − 10)6 5 ⇒ C=− 85.96 . (500)6 and, according to (2.11), the concentration of salt is given by A(t) 1 85.96 =− · 510 − t 5 510 − t 6 510 − t 500 6 . 20 minutes after the leak develops, that is, when t = 30, the concentration will be 1 85.96 − · 5 510 − 30 37. We are solving the equation πt dx + 2x = 1 − cos dt 12 , x(0) = 10. 510 − 30 500 ≈ 0.0598 kg/L . This is a linear problem with dependent variable x and independent variable t so that P (t) = 2. Therefore, to solve this equation we first must find the integrating factor µ(t). µ(t) = exp Multiplying the equation by this factor yields e2t ⇒ 56 dx + 2xe2t = e2t 1 − cos dt xe2t = e2t dt − πt 12 e2t cos = e2t − e2t cos πt 12 πt 12 e2t cos πt dt. 12 2 dt = e2t . 1 dt = e2t − 2 ...
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