- #1
TranscendArcu
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Homework Statement
The Attempt at a Solution
I let Q be the concentration of salt in the tank at any time t. Thus, the differential equation should be [itex]\frac{dQ}{dt} = (\frac{1}{10})(2) - \frac{Q(1)}{100 + t} = \frac{1}{5} - \frac{Q}{100 + t}[/itex]. Rearranging terms,
[itex]\frac{dQ}{dt} + \frac{Q}{100 + t} = \frac{1}{5}[/itex]. We construct an integrating factor: [itex]θ(t) = e^{\int \frac{1}{100 + t} dt} = e^{ln(100 + t)} = 100 + t[/itex]. We introduce the integrating factor,
[itex](Q(100 + t))' = 20 + \frac{t}{5}[/itex] and integrate both sides to find [itex]Q(100 + t) = 20t + \frac{t^2}{10} + C[/itex]. Thus,
[itex]Q = \frac{20t + \frac{t^2}{10} + C}{100 + t}[/itex]. Given that [itex]Q(0) = 0[/itex], we can solve for C: [itex]Q(0) = 0 = \frac{20(0) + \frac{(0)^2}{10} + C}{100 + (0)} = \frac{C}{100}[/itex] which implies that C must equal zero. Thus,
[itex]Q(100) = \frac{20t + \frac{t^2}{10}}{100 + t} = \frac{2000 + 1000}{200}[/itex]
Have I done this correctly? I suppose my answer is more like the concentration after 100 minutes, but I'm worried because it seems like I don't have much simplifying to do, as the problem hints that I should. Suggestions?