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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 Upon integrating and exponentiating, we find that ￿t ln |µ| = p(τ ) dτ + C1 =⇒ µ(t) = C2 exp ￿￿ t p(τ ) dτ ￿ where C2 = ±eC1 is some constant. This is the general solution to (ii). Since we want µ = µ(t) to satisfy (i), we want C2 ￿= 0. In fact, we will choose C2 = 1 – although this is not necessary. The function µ = µ(t) is called an integrating factor. We conclude the general solution of the original differential equation y ￿ = G(t, y ) in terms of G(t, y ) = g (t) − p(t) y must be ￿￿ t ￿ ￿￿ t ￿ 1 y (t) = µ(τ ) g (τ ) dτ + C where µ(t) = exp p(τ ) dτ . µ(t) Example 1. We revisit the differential equation y ￿ = −b + a y for constants a and b. In this case, y ￿ + p(t) y = g (t) where p(t) = −a and g (t) = −b are constant functions. The integrating factor would be ￿￿ t ￿ ￿￿ t ￿ µ(t) = exp p(τ ) dτ = exp −a dτ = exp [−a t] = e−at . Hence the general solution to the differential equation y ￿ = −b + a y must be ￿￿ t ￿ ￿￿ t ￿ ￿ ￿ 1 b at −at at b −at µ(τ ) g (τ ) dτ + C = e −b e dτ + C = e e + C = + C eat . y (t) = µ(t) a a Example 2. Say that we wish to solve the initial value problem t y ￿ + 2 y = 4 t2 , y (1) = 2. First we find the general solution to the differential equation, then we determine the constants of integration from the initial condition. Divide both sides of the equation by t so that the differential equation is in the form dy + p(t) y = g (t) dt The integrating factor is µ(t) = exp ￿￿ t where ￿ ￿￿ p(τ ) dτ = exp p(t) = t 2 , t g (t) = 4 t. ￿ 2 dτ = exp [2 ln t] = t2 . τ Multiply both sides of the differential equation y ￿ + p(t) y = g (t) by µ(t) = t2 : dy + 2 t y = 4 t3 . dt The left-hand side is the derivative of µ(t) y = t2 y , so we have ￿ d￿2 C t y (t) = 4 t3 =⇒ t2 y (t) = t4 + C =⇒ y (t) = t2 + 2 . dt t This is the general solution to the differential equation. For the initial condition, set t = 1: 1 1 + C = y (1) = 2 =⇒ C=1 =⇒ y (t) = t2 + 2 . t This is the solution to the initial value problem. t2 First Order Nonlinear Equations Definitions. Every first order differential equation can be expressed in the form dy = G(t, y ). dt Recall that we say this equation is linear if G(t, y ) = g (t) − p(t) y for some functions p = p(t) and g = g (t). We say that this equation is nonlinear otherwise. We say this first order differential equation is separable if we have the factorization G(t, y ) = Y (y ) T (t) in terms of functions Y = Y (y ) and T = T (t). ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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