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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 First Order Linear Equations Definitions. So far we have considered initial value problems in the form dy = −b + a y, y (0) = y0 dt where a and b are constants. We showed that the solution is b b y (t) = + C eat where C = y0 − . a a In general, an ordinary differential equation is said to be a first order equation if it is a relation in the form F (t, y, y ￿ ) = 0. Upon expressing the derivative y ￿ in terms of the other variables, we find an expression in the form dy = G(t, y ) dt for some function G(t, y ). We say that a function y = y (t) is a solution if it satisfies this equation. If y (t0 ) = y0 is an initial value, we think of the graph of y = y (t) as passing through the point (t0 , y0 ). We say that this first order equation is linear if G(t, y ) = g (t) − p(t) y for some functions g (t) and p(t). That is, if the differential equation is in the form dy + p(t) y = g (t). dt Integrating Factors. Consider the first order linear differential equation dy + p(t) y = g (t). dt Choose a function µ = µ(t). Multiply the first order differential equation above by this function: dy + µ(t) p(t) y = µ(t) g (t). dt Say that we can choose a function µ = µ(t) such that i. µ(t) is not identically zero, and ii. µ￿ (t) = p(t) µ(t). Then we would have dy dy dµ d µ(t) + µ(t) p(t) y = µ(t) · + · y (t) = [µ(t) y (t)] dt dt dt dt This gives the differential equation ￿t d [µ(t) y (t)] = µ(t) g (t) =⇒ µ(t) y (t) = µ(τ ) g (τ ) dτ + C dt µ(t) for some constant C . (We use this notation to remind ourselves that the indefinite integral of a function is really a function of t, not the dummy variable of integration τ .) Before we continue, we study conditions (i) and (ii) in more detail. If µ = µ(t) is not identically zero, we can divide by it: dµ 1 dµ = p(t) µ =⇒ = p(t). dt µ dt The left-hand side is a derivative: 1 dµ d d = ln |µ| =⇒ ln |µ| = p(t). µ dt dt dt 1 ...
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