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Homework Help: Basic Differential Eqn - initial value problem

  1. Jan 11, 2006 #1


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    The equation starts with dy/dt = ay - b, and initial condition y(0) = y_o and y_o is an arbitrary value. The book says this can be rewritten as:
    (dy/dt)/[y - (b/a)] = a. But I don't see how do make that step algebraically. How can the original be rewritten like that?

    A problem at the end of the chapter appears to be designed to use the above equation.
    (dy/dt) = -y + 5

    Am I correct in assuming I should rewrite it as this?
    (dy/dt)/[y - 5/(-1)] = -1

    Then follow the steps to solve that? Beginning with:
    (ln[y - (5/(-1)]) = -1t + C
  2. jcsd
  3. Jan 11, 2006 #2
    For the first step, factor an a out of the right hand side then divide both sides by (y-b/a).
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