5389-Chap2 - Chapter 2 First Order Dierential Equations...

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Chapter 2 First Order Differential Equations Introduction Any frst 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, and y ± must appear explicitly in the expression F . Our study oF frst 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 ) . ( * ) 2.1 Linear Differential Equations A frst 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 ). The precise defnition oF a linear equation that we will use is: The frst order differential equation y ± = f ( x, y )i sa linear equation iF it can be written in the Form y ± + p ( x ) y = q ( x ) (1) 15
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where p and q are continuous functions on some interval I . Differential equations that are not linear are called nonlinear equations. SOLUTION METHOD Step 1. Identify: If possible, write the equation in the form (1): y ± + p ( x ) y = q ( x ). Step 2. Calculate h ( x )= ± p ( x ) dx (omitting the constant of integration) and form e h ( x ) . Step 3. Multiply the equation by e h ( x ) to obtain e h ( x ) y ± + e h ( x ) p ( x ) y = e h ( x ) q ( x ) . Verify that the left side of this equation is ² e h ( x ) y ³ ± . Thus we have ´ e h ( x ) y µ ± = e h ( x ) q ( x ) . Step 4. The equation in Step 3 implies that e h ( x ) y = ± e h ( x ) q ( x ) dx + C and y = e - h ( x ) ¶± e h ( x ) q ( x ) dx + C · = e - h ( x ) ± e h ( x ) q ( x ) dx + Ce - h ( x ) . Therefore, the general solution of (1) is: y = e - h ( x ) ± e h ( x ) q ( x ) dx + - h ( x ) . (2) INTEGRATING FACTOR The key step in solving y ± + p ( x ) y = q ( x ) is multiplication by e h ( x ) where h ( x ¸ p ( x ) dx . It is multiplication by this factor, called an integrating factor, that enables us to write the left side of the equation as a derivative (the derivative of the product e h ( x ) y ) from which we get the general solution in Step 4. ± EXISTENCE AND UNIQUENESS Obviously solutions of Frst order linear equations exist. It follows from Steps (3) and (4) that the general solution (2) represents all solutions of the equation (1). As you will see, if an initial condition is speciFed, then the constant C will be uniquely determined. Thus, a Frst order, linear, initial-value problem will have a unique solution. 16
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Example 1. Find the general solution of y ± +2 xy = x. SOLUTION (1) The equation is linear; it is already in the form (1); p ( x )=2 x, q ( x )= x , continuous functions on ( -∞ , ). (2) Calculate: h ( x ± 2 xdx = x 2 and e h ( x ) = e x 2 . (3) Multiply by e x 2 : e x 2 y ± xe x 2 y = x 2 ² e x 2 y ³ ± = x 2 (verify this) (4) Integrate: e x 2 y = ± x 2 dx = 1 2 e x 2 + C and y = e - x 2 ´ 1 2 e x 2 + C µ = 1 2 + Ce - x 2 .
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5389-Chap2 - Chapter 2 First Order Dierential Equations...

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