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Ordinary and Partial Diﬀerential Equations Lecture 2. First-order diﬀerential equations - Part 1. Separable equations. Linear equations. Bibliography: Nagle, Snaﬀ, Sieder - ”Fundamentals of DEs and BVPs” (6th ed.) - sections 2.1, 2.2, 2.3 INTRODUCTION: Motion of a falling body Problem: An object falls through the air toward Earth. Assuming that the only forces acting on the object are gravity and air resistance, determine the velocity of the object as a function of time. Newton’s second law can be applied to the falling object, leading to the diﬀerential equation: dv =F dt where v = v(t) is the velocity of the body at the moment of time t, m is the mass of the body, and F is the dv total force acting on the body. The ﬁrst-order derivative of the velocity represents the acceleration. dt Gravity can be expressed by mg, where g is the gravitational acceleration. Air resistance can be reasonably represented by −bv, where b is a positive constant depending on the density of the air and the shape of the object, and v is the velocity of the object. We use the negative sign because air resistance is a force that opposes the motion. Hence, our diﬀerential equation becomes: m dv = mg − bv (1) dt This is a ﬁrst-order linear diﬀerential equation, with the dependent variable v = v(t). We solve this equation using a technique called separation of variables. Isolating the variables v and t on opposite sides of the equation, we obtain: ∫ ∫ dv dt dv dt 1 t mg = ⇒ = ⇒ − ln |mg − bv| = +c ⇒ v = − C · e−bt/m mg − bv m mg − bv m b m b mg Associating the initial condition v(0) = v0 to the diﬀerential equation (1), we ﬁnd C = v0 − . b ( ) mg mg The solution of this IVP is: v(t) = − v0 − · e−bt/m . b b m SEPARABLE EQUATIONS ♣ A separable diﬀerential equation is and equation of the form: dy = g(x)p(y) (2) dx where the function g depends only on the independent variable x and the function p depends only on the dependent variable y. Method for solving separable equations: • separate the variables on opposite sides of the equation: 1 , and so: h(y)dy = g(x)dx p(y) ∫ ∫ • integrate on both sides: h(y)dy = g(x)dx 1 dy = g(x)dx p(y) • denote h(y) = ⇒ H(y) = G(x) + c where H and G are primitives of the functions h and g, respectively. • the equation H(y) = G(x) + c represents an implicit solution for the separable equation (2). Caution: Constant functions y ≡ c such that p(c) = 0 are also solutions to (2), which may or may not be found by this method. c 2011/2012 Eva Kaslik - West University of Timişoara, Romania ⃝ LINEAR EQUATIONS A ﬁrst-order linear diﬀerential equation is of the form a1 (x) dy + a0 (x)y = b(x) dx (3) where the function a0 (x), a1 (x) and b(x) do not depend on the dependent variable y. Method for solving linear equations: • write eq. (3) in the standard form: dy a0 (x) b(x) + P (x)y = Q(x), where P (x) = and Q(x) = dx a1 (x) a1 (x) • calculate the integrating factor µ(x) = e • notice that ∫ P (x)dx ∫ dµ = P (x)e P (x)dx = P (x)µ(x) dx • multiply the equation by µ(x) to obtain µ(x) dy + P (x)µ(x)y = Q(x)µ(x) dx ⇔ µ(x) dy dµ + y = Q(x)µ(x) dx dx ⇔ d [µ(x)y] = Q(x)µ(x) dx ∫ • integrate the last equation: µ(x)y = Q(x)µ(x)dx + C • obtain the explicit solution: (∫ ) (∫ ) [ ∫ ] 1 y(x) = Q(x)µ(x)dx + C = Q(x)µ(x)dx + C exp − P (x)dx µ(x) (4) Existence and Uniqueness of the Solution Suppose P and Q and are continuous on an interval (a, b) that contains the point x0 . Then for any choice of initial value y0 , there exists a unique solution on (a, b) to the initial value problem dy + P (x)y = Q(x) dx (∫ The solution is: ) x y(x) = ; Q(s)µ(s)ds + y0 x0 y(x0 ) = y0 [ ∫ exp − P (s)ds x0 2 ] x