141Bquiz7csols

# 141Bquiz7csols - 1 MATH 141B QUIZ#7 Solutions Date Tues 1 V...

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1 MATH 141B QUIZ#7 Solutions Date: Tues March 18, 2008 1. V = 100 and F = 25 , so F/V = 1 / 4 . Since I = c · F = (0 . 1)(25) = 2 . 5 , the diferential equation is y - 1 4 y + 2 . 5. In standard Form, we get y + 1 4 y = 5 2 . Now, F ( t ) = i 1 4 dt = 1 4 t, so the integrating Factor is e t/ 4 . Multiplying by G gives, e t/ 4 y + 1 4 e t/ 4 y = 5 2 e t/ 4 . By the product rule For diferentiation, this becomes ( ye t/ 4 ) = 5 2 e t/ 4 . Integrating, and we get ye t/ 4 = 10 e t/ 4 + C. We don’t need to ±nd C since it will vanish in the limit. Multiplying by e t/ 4 gives y ( t ) = 10 + Ce ( t/ 4) . ²inally, y = lim t →∞ y ( t ) = 10 . ThereFore, A) is the answer. 2. In standard Form, the diferential equation is s + ( ln 2 30 ) s = 1 2 . Now, F ( t ) = i ln 2 30 dt = ln 2 30 t. ThereFore, the integrating Factor is G ( t ) = e (ln 2 / 30) t . Multiplying the D.E. by G ( t ) gives [ s · e (ln 2 / 30) t ] = 1 2 e (ln 2 / 30)
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