DE c and m chapter 1 sc

CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS SECTION 1.1 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS The main purpose of Section 1.1 is simply to introduce the basic notation and

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CHAPTER 1

FIRST-ORDER DIFFERENTIAL EQUATIONS

SECTION 1.1

DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS

The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of dif-ferential equations, and to show the student what is meant by a solution of a difdif-ferential equation Also, the use of differential equations in the mathematical modeling of real-world phenomena is outlined

Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the given differential equations We include here just some typical examples of such verifications

3 If y1cos 2x and y2 sin 2x, then y1  2sin 2x y2 2 cos 2x, so

1 4 cos 2 4 1

y  x  y and y2  4 sin 2x 4y2 Thus y14y1 and 0 y2 4y2  0

1

x

y e and 3

2

x

y e , then 3

1 3 x

2 3 x

y   e , so 3

1 9 x 9 1

y e  y and 3

2 9 x 9 2

y  e  y

y e e , then x x

y  e e , so y  y e xex  e xex  2ex Thus

2 x

y  y e

1

x

y e and 2

2

x

y x e , then 2

1 2 x

y   e , 2

1 4 x

y e , 2 2

2 x 2 x

y e  x e , and

2 4 x 4 x

y   e  x e Hence

 2   2   2

1 4 1 4 1 4 x 4 2 x 4 x 0

y y y  e   e  e 

and

2 4 2 4 2 4 x 4 x 4 x 2 x 4 x 0

y y y   e  x e  e  x e  x e 

8 If y1cosxcos 2x and y2 sinxcos 2x, then y1  sinx2 sin 2 ,x

1 cos 4 cos 2 ,

y  x x y2 cosx2sin 2x, and y2  sinx4 cos 2 x Hence

yy   x x  x x  x

and

2 2 sin 4 cos 2 sin cos 2 3cos 2

y y   x x  x x  x

2 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS

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11 If yy1 x2, then y  2x3 and y 6x4, so

x y x y y  x x  x  x  x 

2 ln

yy x x, then y x32x3lnx and y  5x46x4lnx, so

x y x y y x x x x x x x x x x

x x x x x x

13 Substitution of rx

ye into 3y 2y gives the equation 3 rx 2 rx

r e  e , which simplifies

to 3r2 Thus r2 / 3

14 Substitution of rx

ye into 4 y  y gives the equation 4 2 rx rx

r e  e , which simplifies to 2

4r  Thus 1 r 1 / 2

15 Substitution of ye rx into y y 2y  0 gives the equation r e2 rxr e rx2e rx  , 0

which simplifies to r2  r 2 (r2)(r1)  0 Thus r 2 or r1

16 Substitution of rx

ye into 3y3y4y0 gives the equation 3 2 rx 3 rx 4 rx 0

r e  r e  e  , which simplifies to 3r2   The quadratic formula then gives the solutions 3r 4 0

The verifications of the suggested solutions in Problems 17-26 are similar to those in Problems

1-12 We illustrate the determination of the value of C only in some typical cases However, we

illustrate typical solution curves for each of these problems