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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 3 Separable Equations. We introduce a trick to solve separable differential equations. Divide both sides by the function Y = Y (y ) so that all of the y terms are to one side: dy 1 dy = Y (y ) T (t) =⇒ = T (t). dt Y (y ) dt Now compute the antiderivatives ￿t d f (t) = T (t) f1 (t) = T (τ ) dτ 1 dt =⇒ ￿y d 1 f2 (y ) = 1 dy f2 (y ) = dσ dt Y (y ) dt Y (σ ) We see that the function f (t, y ) = f1 (t) − f2 (y ) satisfies d d d 1 dy f (t, y ) = f1 (t) − f2 (y ) = T (t) − = 0, dt dt dt Y (y ) dt so that f (t, y ) must be a constant. We conclude that the general solution to the differential equation y ￿ = G(t, y ) in terms of G(t, y ) = Y (y ) T (t) must be ￿t ￿y 1 f (t, y ) = C where f (t, y ) = T (τ ) dτ − dσ. Y (σ ) This is an implicit solution to the differential equation y ￿ = G(t, y ) because we have not explicitly written y = y (t) as a function of time. We also call these integral curves because they give relations between y and t. The constant C can be found from the initial conditions. Example. Once again we revisit the differential equation y ￿ = −b + a y for constants a and b. In this case, y ￿ = Y (y ) T (t) where we choose Y (y ) = y − (b/a) and T (t) = a. The antiderivatives are ￿t ￿t f1 (t) = T (τ ) dτ = a dτ = a t ￿ ￿ ￿y ￿y ￿ ￿ 1 dσ ￿y − b ￿ . f2 (y ) = dσ = = ln ￿ Y (σ ) σ − (b/a) a￿ Hence the integral curves are ￿ ￿ ￿ ￿ b ￿y − b ￿ = C a t − ln ￿ =⇒ y = + C1 eat a￿ a in terms of the constant C1 = ±e−C . Hence the integral curves is just the general solution. ...
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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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