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 May 31st, 2015, 05:19 PM #1 Newbie   Joined: May 2015 From: Waterloo, Canada Posts: 2 Thanks: 0 Math Focus: Calculus A math problem that involves first order ODE Hi all, I came across this problem today. This is my attempted solution. As you can see, I stumbled upon an equation that is in implicit form. I found C by substituting in the initial values given, however unable to find v when y = 0 because it is in implicit form. Am I doing this wrong? June 2nd, 2015, 01:09 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 let the downward direction be positive with $y_0 = 0$, $v_0=0$, and $y_f=30$ $\displaystyle m \frac{dv}{dt} = mg - kv$ $\displaystyle \frac{dv}{dt} = g - \frac{k}{m}v$ $\displaystyle \frac{dv}{dt} = g - 0.8v$ $\displaystyle \frac{dv}{g-0.8v} = dt$ $\displaystyle \ln|g-0.8v| = -0.8t + C$ $\displaystyle g-0.8v = Ae^{-0.8t}$ $\displaystyle v(0) = 0 \implies A = g$ ... $\displaystyle 0.8v = g(1 - e^{-0.8t})$ $\displaystyle v = \frac{5g}{4}(1 - e^{-0.8t})$ ... note terminal velocity, $\displaystyle v_T = \frac{5g}{4} \, m/s$ $\displaystyle \frac{dy}{dt} = \frac{5g}{4}(1 - e^{-0.8t})$ $\displaystyle y = \frac{5g}{4}(t + 1.25e^{-0.8t} + C)$ $\displaystyle y(0) = 0$ $\displaystyle 0 = \frac{5g}{4}(1.25 + C) \implies C = -1.25$ $\displaystyle y = \frac{5g}{4}[t + 1.25(e^{-0.8t} - 1)]$ $\displaystyle 30 = \frac{5g}{4}[t + 1.25(e^{-0.8t} - 1)]$ using $\displaystyle g = 9.8 \, m/s^2$ and solving with a calculator ... $\displaystyle t \approx 3.63 \, sec$ $\displaystyle v(3.63) \approx 11.82 \, m/s$ Thanks from Tu TN Tags calculus, differential equation, first order, involves, math, ode, order, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Monox D. I-Fly Elementary Math 19 May 24th, 2016 01:35 AM E01 Academic Guidance 7 February 9th, 2014 08:25 AM fairyrak Real Analysis 5 December 30th, 2011 05:39 PM jackie Real Analysis 1 February 25th, 2008 10:30 AM

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