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2 MA 36600 LECTURE NOTES: FRIDAY, JANUARY 16 The initial value problem is m dv dt = mg γv ° Diferential Equation v (0) = v 0 ° Initial Condition This is a little more difficult to solve, but not impossible. The problem is we cannot simply integrate both sides because the right-hand side involves a Function oF v . However, we will try and move all oF the v ’s to one side. The diferential equation is dv dt = g γ m v = γ m ± v + mg γ ² = 1 v + mg γ dv dt = γ m . Observe that For any constant c we have d dt ln | y + c | = 1 y + c dy dt by the Chain Rule. This means we have the expression d dt ln ³ ³ ³ ³ v + mg γ ³ ³ ³ ³ = 1 v + mg γ dv dt = γ m = ln ³ ³ ³ ³ v + mg γ ³ ³ ³ ³ = γ m t + C For some constant C . Upon exponentiating both sides, we ±nd that ³ ³ ³ ³ v + mg γ ³ ³ ³ ³ = e ( γ/m ) t + C = e C · e ( γ/m ) t = v + mg γ = C 1 e ( γ/m ) t where C 1 = ± e C is another constant. ²or the initial condition, we set t = 0: C 1 = v 0 + mg γ = v ( t )= mg γ + ± v 0 + mg γ ² e (
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