Chapter 2 Slides

# Chapter 2 Slides - FIRST ORDER EQUATIONS F (x, y, y ) = 0...

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FIRST ORDER EQUATIONS F ( x,y,y ± )=0 Basic assumption: This equation can be solved for y ± ; that is, the equation can be written in the form y ± = f ( x, y ) (1) 1

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I. First Order Linear Equations Equation (1) is a linear equation if f has the form f ( x,y )= P ( x ) y + q ( x ) where P and q are continuous functions on some interval I . Thus y ± = P ( x ) y + q ( x ) 2
Standard form: The standard form for a Frst order linear equation is: y ± + p ( x ) y = q ( x ) where p and q are continuous functions on the interval I 3

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Solution Method: Step 1. Establish that the equa- tion is linear and write it in standard form y ± + p ( x ) y = q ( x ) . Step 2. Multiply by e ± p ( x ) dx : e ± p ( x ) dx y ± + p ( x ) e ± p ( x ) dx y = q ( x ) e ± p ( x ) dx ² e ± p ( x ) dx y ³ ± = q ( x ) e ± p ( x ) dx 4
Step 3. Integrate: e ± p ( x ) dx y = ² q ( x ) e ± p ( x ) dx dx + C. Step 4. Solve for y : y = e - ± p ( x ) dx ² q ( x ) e ± p ( x ) dx dx + Ce - ± p ( x ) dx . This is the general solution of the equation. 5

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Examples: 1. Find the general solution: y ± +2 xy =4 x Answer: y =2+ Ce - x 2 2. Find the general solution: xy ± +3 y = ln x x Answer: y = ln x 2 x - 1 4 x + C x 3 6
3. Solve the initial-value problem: y ± +(cot x ) y = 2cos x, y ( π/ 2) = 3 Answer: y = 2 sin x - cos 2 x sin x 7

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Important Special Case y ± + p ( x ) y =0 . General solution: y = Ce - ± p ( x ) dx . Since e - ± p ( x ) dx ± = 0 for all x ,i t follows that: 1. If y ( a ) = 0 for some a , then C = 0 and y 0. 2. If y ( a ) ± = 0 for some a , then C ± = 0 and y ( x ) ± = 0 for all x . 8
The term “linear:” L [ y ]= y ± + p ( x ) y is a linear operator: L [ y 1 + y 2 L [ y 1 ]+ L [ y 2 ] and L [ cy cL [ y ] . 9

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A related equation: Transforma- tion into a linear equation: Bernoulli equations An equation of the form y ± + p ( x ) y = q ( x ) y k ,k ± =0 , 1 is called a Bernoulli equation . 10
The change of variable v = y 1 - k transforms a Bernoulli equation into the linear equation v ± +(1 - k ) p ( x ) v =(1 - k ) q ( x ) . 11

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Examples: 1. Find the general solution: y ± - 4 y =2 e x y Answer: y =( Ce 2 x - e x ) 2 2. Find the general solution: xy ± + y =3 x 3 y 2 Answer: y 2 = - 3 2 x 3 + Cx 12
II. Separable Equations Equation (1) is a separable equa- tion if f has the form f ( x,y )= p ( x ) h ( y ) where p and h are continuous functions. Thus y ± = p ( x ) h ( y ) 13

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Solution Method Step 1. Establish that the equa- tion is separable. Step 2. Divide both sides by h ( y ) to “separate” the variables. 1 h ( y ) y ± = p ( x )o r q ( y ) y ± = p ( x ) which, in diﬀerential form, is: q ( y ) dy = p ( x ) dx.
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## This note was uploaded on 02/09/2011 for the course MATH 3321 taught by Professor Morgan during the Fall '08 term at University of Houston.

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Chapter 2 Slides - FIRST ORDER EQUATIONS F (x, y, y ) = 0...

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