[Dennis.G.zill] a.first.course.in.Differential.equations.9th.ed Solutions Page 22

Exercises 1.3

Differential Equations as Mathematical Models

20 . From Pr oblem 19. w ithout a damping force, force, the differ differ ential equation is is rnd2x rnd2x /dt 2 =  —kx .  With a damping force proportional to velocity, the differential equation becomes d2x  d2x  , ,.dx  rn — rn —7T =  —kx —p — dt 2 ’ dt

d2x d2x dt/

m ~37> ~37> +

or

A x  dt

f t—

+

k'x

=

0.

2 1 . From g — k / R 2  2  we find k  = gli2.  gli2.  Using a = d2r/d,t 2  and the fact that the positive direction is

upward wc get

___k 

cfr= _

dt2

a

g R2

d2r

or

r2

gR?

r2 *5+1^ =a

i 1f

-----------  

—I—

----

22. T he gr av itational force force on m is F = —kMr m /r2. /r2.  Since Mr = 4irSr ^/3  and M = 47r5 i ?3/3   wc have M r = i : iM iM j R 3 and F = -k 

Mrm

= - k 

- k

=

m.M

r.

R*

Now from F = rria rria = d?7'/dt ?7'/dt2 2 we have d-r d- r , rnM m —w = —w  =  —k   —k — ——r- r dt 2 R*

or

d~r —=• dt 2

kM

r.

dA  23. T he differential differe ntial equation is —- = k(M —A). dt dA  24. T he differential differe ntial equation is = ki(M —A) —k^A. dt 25. T he differential differe ntial equation is x'(t) = r —kx(t ) whore k  > 0 . 26. By the the Pv thagorcan T heorem heorem the the slope slope of the tangent line is is  y — ,— 27. We see see from fr om the figure tha t 29 + a = tt. tt. T hus  V  2 tan ta n 9  —- = tan ta n a   = tan( 7r  —29) = —tan 2$ =   - - - - - - . 1  —tan 2 9 - x  v ; Sincc the slope of the tangent line is ?/

- tan 0 we have y have  y /x = 2y '/[ '/[ l —{yr)2] —{yr)2]

or  y —y(y —y (y ')‘2 = 2xy', xy',  which is the quadratic equation  y(y ')'2 ')'2 + 2xy' —y —i) in  y '. Using the quadr quadratic atic formul for mula, a, we get f

—2 x ± ^4 x 2 + 4y 2

—x ± i j x 2 + y 2

2y

y 

 V

Since dy/dx > 0 , the differential equation is dy _ - x   + + ^ x  2 +  y 2 dx 

or

 y 

dy  dx 

.lie differential equation is dP/dt = kP,  kP,   so from Problem 41 in Exercises 1.1, P ~ eki, and a e- param paramcter cter family of solutions solutions is P = cekt. 19