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I don't understand this problem: Suppose y(t) = 9e - 5t is a solution of the initial value problem y' + ky = 0 what is the value of k?
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\( y' + ky = 0\) is separable. As follows: \( y' = - ky \) \(\dfrac{ y'}{y} = - k \) So integrate: \(\int \dfrac{ dy}{y} = - k \int dx\) \(\implies \ln y = - k x + C\) Take logs both sides: \(y = e^{- k x + C} = e^{- k x } e^C = A e^{- kx}\) We've just played with the integration constant. Then pattern match back to your solution: \(y = 9e^{ - 5t}\)
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