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# S21 - CHAPTER 2 First Order Differential Equations...

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CHAPTER 2 First Order Differential Equations Introduction It follows from the discussion in Section 1.3 that any first order differential equation can be written as F ( x, y, y ) = 0 by moving all nonzero terms to the left hand side of the equation. To be a first order differential equation, y must appear explicitly in the expression F . Our study of first order differential equations requires an additional assumption, namely that the equation can be solved for y . This means that we can write the equation in the form y = f ( x, y ) . ( * ) The study of first order differential equation is somewhat like the treatment of techniques of integration in Calculus II. There we had a variety of ”integration methods” (integration by parts, trigonometric substitutions, partial fraction decomposition, etc.) which were applied according to the form of the integrand. In this chapter we will learn some solution methods for ( * ) which will correspond to the form of the function f . 2.1 Linear Differential Equations A first order differential equation y = f ( x, y ) is a linear equation if the function f is a “linear expression” in y . That is, the equation is linear if the function f has the form f ( x, y ) = P ( x ) y + q ( x ) . (c.f. the linear function y = mx + b .) The solution method for linear equations is based on writing the equation as y - P ( x ) y = q ( x ) which is the same as y + p ( x ) y = q ( x ) where p ( x ) = - P ( x ). We will focus on the latter form. The precise definition of a linear equation that we will use is: FIRST ORDER LINEAR DIFFERENTIAL EQUATION The first order differential equa- tion y = f ( x, y ) is a linear equation if it can be written in the form y + p ( x ) y = q ( x ) (1) where p and q are continuous functions on some interval I . Differential equations that are not linear are called nonlinear differential equations. Example 1. Some examples of linear equations are: 19

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(a) y = 2 xy . Written in the form (1): y - 2 xy = 0 , where p ( x ) = - 2 x, q ( x ) = 0 are continuous functions on ( -∞ , ). (b) y + 2 xy = x is already in the form (1): p ( x ) = 2 x, q ( x ) = x continuous on ( -∞ , ). (c) xy + 2 y = e 3 x x . Written in the form (1): y + 2 x y = e 3 x x 2 , where p ( x ) = 2 /x, q ( x ) = e 3 x /x 2 are continuous functions on any interval that does not contain 0. For example, (0 , ) or ( -∞ , 0) each satisfy this requirement. (d) (1 - x 2 ) y + 2 xy = x 2 - 1. Written in the form (1): y + 2 x 1 - x 2 y = - 1 , where p ( x ) = 2 x 1 - x 2 , q ( x ) = - 1 are continuous on any interval that does not contain 1 and - 1. For example, ( - 1 , 1) or (1 , ) each satisfy this requirement. Solution Method for First Order Linear Equations Step 1. Identify: Can you write the equation in the form (1): y + p ( x ) y = q ( x ) ? If yes, do so. Step 2. Calculate h ( x ) = p ( x ) dx (omitting the constant of integration) and form e h ( x ) .
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S21 - CHAPTER 2 First Order Differential Equations...

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