AEM_3e_Chapter_01

AEM_3e_Chapter_01 - Part I Ordinary Differential Equations...

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Part I Ordinary Differential Equations 1 1 Introduction to Differential Equations EXERCISES 1.1 Definitions and Terminology 3. Fourth order; linear 6. Second order; nonlinear because of R 2 9. Writing the di±erential equation in the form x ( dy/dx )+ y 2 = 1, we see that it is nonlinear in y because of y 2 . However, writing it in the form ( y 2 1)( dx/dy x = 0, we see that it is linear in x . 12. From y = 6 5 6 5 e 20 t we obtain dy/dt =24 e 20 t , so that dy dt +20 y e 20 t µ 6 5 6 5 e 20 t . 15. The domain of the function, found by solving x +2 0, is [ 2 , ). From y 0 =1+2( x +2) 1 / 2 we have ( y x ) y 0 =( y x )[1 + (2( x 1 / 2 ] = y x +2( y x )( x 1 / 2 = y x +2[ x +4( x 1 / 2 x ]( x 1 / 2 = y x +8( x 1 / 2 ( x 1 / 2 = y x +8 . An interval of de²nition for the solution of the di±erential equation is ( 2 , ) because y 0 is not de²ned at x = 2. 18. The function is y =1 / 1 sin x , whose domain is obtained from 1 sin x 6 = 0 or sin x 6 = 1. Thus, the domain is { x ¯ ¯ x 6 = π/ 2+2 } . From y 0 = 1 2 (1 sin x ) 3 / 2 ( cos x )wehave 2 y 0 =(1 sin x ) 3 / 2 cos x = [(1 sin x ) 1 / 2 ] 3 cos x = y 3 cos x. An interval of de²nition for the solution of the di±erential equation is ( 2 , 5 2). Another one is (5 2 , 9 2), and so on. 21. Di±erentiating P = c 1 e t / (1 + c 1 e t ) we obtain dP dt = (1 + c 1 e t ) c 1 e t c 1 e t · c 1 e t (1 + c 1 e t ) 2 = c 1 e t 1+ c 1 e t [(1 + c 1 e t ) c 1 e t ] c 1 e t = c 1 e t c 1 e t · 1 c 1 e t c 1 e t ¸ = P (1 P ) . 1
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1.1 Defnitions and Terminology 24. From y = c 1 x 1 + c 2 x + c 3 x ln x +4 x 2 we obtain dy dx = c 1 x 2 + c 2 + c 3 + c 3 ln x +8 x, d 2 y dx 2 =2 c 1 x 3 + c 3 x 1 , and d 3 y dx 3 = 6 c 1 x 4 c 3 x 2 , so that x 3 d 3 y dx 3 +2 x 2 d 2 y dx 2 x dy dx + y =( 6 c 1 c 1 + c 1 + c 1 ) x 1 +( c 3 c 3 c 2 c 3 + c 2 ) x c 3 + c 3 ) x ln x + (16 8+4) x 2 =12 x 2 . 27. (a) From y = e mx we obtain y 0 = me mx . Then y 0 y = 0 implies me mx e mx m +2) e mx =0 . Since e mx > 0 for all x , m = 2. Thus y = e 2 x is a solution. (b) From y = e mx we obtain y 0 = me mx and y 0 = m 2 e mx . Then y 0 5 y 0 +6 y = 0 implies m 2 e mx 5 me mx e mx m 2)( m 3) e mx . Since e mx > 0 for all x , m = 2 and m =3.Thus y = e 2 x and y = e 3 x are solutions. 30. We substitute c into the di±erential equation and use y 0 = y 0 = 0. Solving c 2 c 3=( c + 3)( c 1)=0we see that y = 3 and y = 1 are constant solutions. 33. From x = e 2 t +3 e 6 t and y = e 2 t +5 e 6 t we obtain dx dt = 2 e 2 t +18 e 6 t and dy dt e 2 t +30 e 6 t . Then x y e 2 t e 6 t )+3( e 2 t e 6 t )= 2 e 2 t e 6 t = dx dt and 5 x y =5( e 2 t e 6 t e 2 t e 6 t )=2 e 2 t e 6 t = dy dt .
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This note was uploaded on 01/03/2011 for the course BIS 511 taught by Professor Theodoreholford during the Fall '10 term at Yale.

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AEM_3e_Chapter_01 - Part I Ordinary Differential Equations...

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