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Thomas Calculus 13th [Solutions]

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670 Chapter 9 First-Order Differential Equations<br />

39-42. Example CAS commands:<br />

Mathematica: (assigned functions, step sizes, and values for initial conditions may vary)<br />

Problems 39 - 42 involve use of code from Problems 23 - 34 together with the above code for Eulers method.<br />

9.2 FIRST-ORDER LINEAR EQUATIONS<br />

dy x dy<br />

x<br />

dx dx x x<br />

1. x y e 1 y e<br />

, P( x) 1 , Q( x)<br />

1<br />

P( x) dx ln x<br />

P( x) dx dx ln | x | ln x, x 0 v( x)<br />

e e x<br />

x<br />

x<br />

1 1 1 x<br />

y v( x) Q( x) dx x<br />

e<br />

dx e C<br />

e C<br />

, x 0<br />

v( x)<br />

x x x x<br />

x<br />

x<br />

e<br />

x<br />

x<br />

x dy x dy<br />

x<br />

dx<br />

dx<br />

2. e 2 e y 1 2 y e , P ( x ) 2, Q ( x ) e<br />

P( x) dx 2x<br />

P( x) dx 2dx 2 x v( x)<br />

e e<br />

2 2<br />

y e e dx e dx e C e Ce<br />

1 x x 1 x 1 x x x<br />

2x e<br />

2x e<br />

2x<br />

e<br />

x<br />

3. sin x<br />

dy 3 sin x<br />

2 dx x<br />

3<br />

x<br />

x<br />

3<br />

x<br />

xy 3 y , x 0 y , P( x) 3 , Q( x) sin x<br />

3<br />

3 3 ln 3<br />

3ln ln , 0 ( )<br />

x dx x x x v x e x<br />

3<br />

y 1 x sin dx 1 sin x dx 1 cos x C cos , x 0<br />

x C x<br />

3 3 3 3 3<br />

x x x x x<br />

x<br />

x<br />

4.<br />

tan cos 2 , dy<br />

2<br />

y x y x x<br />

2 2 dx<br />

tan x y cos x ,<br />

2<br />

P( x) tan x, Q( x) cos x<br />

1<br />

sin<br />

1 ln(cos x) 1<br />

tan x dx<br />

x<br />

dx ln | cos x | ln (cos x) , x v( x) e (cos x) sec x<br />

cos x<br />

2 2<br />

1<br />

2<br />

y sec x cos x dx (cos x) cos x dx (cos x) sin x C sin xcos x C cos x<br />

sec x<br />

dy<br />

dy<br />

dx x dx x x x<br />

2<br />

2<br />

ln x 2<br />

5. x 2y 1 1 , x 0 2 y 1 1 , P( x) 2 , Q( x)<br />

1 1<br />

2<br />

2<br />

2 2ln ln , 0 ( )<br />

x dx x x x v x e x<br />

x x x<br />

1 2 1 1 1 1<br />

2<br />

x<br />

1 1 C<br />

2<br />

x x 2<br />

x<br />

2<br />

x<br />

2<br />

x 2 2 x 2<br />

x<br />

y x dx ( x 1) dx x C , x 0<br />

dy<br />

6. (1 x) y y x 1 y , P 1<br />

dx 1 x 1 x<br />

( x ) 1 , Q<br />

x<br />

( x ) 1<br />

1 ln(1 ),<br />

1 x dx x since ln(1 x)<br />

x 0 v( x) e 1<br />

x<br />

x 3/2<br />

3/2<br />

(1 )<br />

2x C<br />

x x x x x x<br />

y 1 x dx 1 x dx 1 2 x C<br />

1 1 1 1 3 3(1 ) 1<br />

x<br />

x<br />

7.<br />

dy 1 1 x/2<br />

y e P( x) 1 , 1 x/2 1<br />

x/2<br />

Q( x) e P( x) dx x v( x)<br />

e<br />

dx 2 2 2<br />

2 2<br />

1 x/2 1 x/2 x/2 1 x/2 1 1 x/2 x/2<br />

y e e dx e dx e x C xe Ce<br />

x/2<br />

e<br />

2 2 2 2<br />

Copyright<br />

2014 Pearson Education, Inc.

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