Differential-Equations-by-Zill-3rd-Edition-Solutions-Manual(ciitbooks.blogspot.com)_text.pdf

1

Introduction to Differential Equations Exercises

1.1

1.

Second-order; linear.

2.

Third-order; nonlinear because of (dy/dx) 4

3.

First-order; nonlinear because of yy'.

4.

First-order; linear.

5.

Fourth-order; linear.

.

6. Second-order; nonlinear because of sin y.

(Sy/dx2^

7.

Second-order; nonbnear because of

8.

2 Second-order; nonlinear because of 1/r

9.

Third-order; linear.

.

.

2

10. First-order, nonlinear because of y x /2

11.

From y = e~

12.

From y = 8 we obtain y

13.

From y =

we

obtain y' 1

e

3*

+

lOe

2*

=

0,

= -\e~ x l 2

6

6

5

5

.

+ y= ~e~ x l 2 + e~^ 2 = + 4(8) = 32.

Then

2y'

we obtain dyjdx

From y = - — -e -20i we obtain

$+

(3e

=

20j,

31

=



y'

x2

m 2 e mx _ 5me mz + fe m* = Since e mx

Since e

-2 < x < is

I

and the

s/T^,

i

Thus y

3.

mi and

rae

=

y"

m 2 e m;E

e .

2x

and y

Then y"

5y'

3)e

=

e

+

ml

3x

=

implies

0.

are solutions.

4- lOj/ 4-

m 2 e mx + IQme™ + 25e mi = (m + 5) 2 e mi = x, m = Thus, y = e 51 a solution. 5.

=

6y

25y

=

implies

0.

is

m~ l

= mx and y" = m(m— l)x m_2 and substituting into the differential equation we obtain m m m(m - l)x - x = (m 2 - m - l) x m = 0. Solving ro 2 - m - 1 = we obtain m = (l ± i/o") /2.

49. Using y'

Thus, two solutions of the differential equation on the interval

50. Using

j/

+ 4y

and y"

.

m(m Thus,

y 51.

—

m=

x~* and y

It is easily y?,

—4,

-

arO+v^)/^

= m(m— l)x m ~ 2 and substituting into the differential equation we obtain = [m(m — 1) + 6m + 4]x m The right side will be zero provided m satisfies

= mx"1-1

x 2 y" + Sxy'

0A (10 -

'

h)

0.

>

where

in

1

A)

0.

—

the x- and ^-directions gives

and

2

d y m— j = — 13.

respectively-

Prom Newton's second law

in the x-direction

we have

2

d x m— dt2

md

2

y

A 15.

To

=

respectively,

and k >

0,

see from the figure that

tan20

=

,

,

Id 11

r-

.

rfi

is

Ady

k(a~x)(j3—x) where a and

rfi

=

20, tan



1

—^ =

2tan0

- tan-^

x

x

L down

-, so

- =

j/

y

—-

2ldx/dy)

1

f 7T

I

L-~~ and i

(dx dy) 2

6

^

~dt

are the given

— — tan — — 9 \2 ax

,

dy lcl

to the x-axis.

x dy — — and y

=

1

.

better understand the problem extend the line

tan?i

dx —dx = — — v dt

,

x'(t)

differential equation

and B,

,

= ~ mg ksm0a = ~ m9 ~ k = ~ mg ~ ~ vTt

-k¥

The

,

we have

In the y-direction

14.

„

—kcos6 — —k

-

2

)

=

amounts of chemicals

Then we

cot

G.

Now

'V y

—

(dx\ „ ( dx\ = +2y \dy ) \dyj

—

L

i.

Exercises 1.2

16.

We

have from Archimedes' principle

upward

It

force of water

on barrel

then follows from Newton's second law that

=

weight of water displaced

=

(62.4)

=

(62.4)7r(s/2) y

x (volume of water displaced) 2

—^-^ =

—15. 6tts

g dt 1 g 17.

=

w

32 and

~ &i

'

m gives

Dividing by

2

15.67rs y.

2

or

i(

dt 2

w iere '

^-f 4 dt

M

the mass of the earth and k\

is

= --4£

of motion with his law of gravitation,

k\M = gR j/B is

,

3

=

w

q wnere

where k

,

=

k\M. The constant k

= gR

2

or k

.

we obtain

a constant of proportionality.

is

is

R is the

gR 2 where ,

y

the earth. This follows from the fact that on the surface of the earth y 2

15.67T8

+

the weight of the barrel in pounds.

is

By combining Newton's second law

m ^f" =

=

=

If t

is

=R

radius of

^" = m9<

so that ^l -

the time at which burnout occurs, then y(0)

the distance from the earth's surface to the rocket at the time of burnout,

= R + yg, where and y'(0) — Vg is

the corresponding velocity at that time. 18. Substituting into the differential equation

we obtain —(mo

—

at)g

=

du — at)— + M—a)

(mo

or

dt

i

(mo

dv

— at)— = i

,

ab

— m^g + agt.

at

19.

By

the Pythagorean

Theorem the

slope of the tangent line

is y'

^

—

.

2

20. (a)

We

have

M

r

=

\l»

M •So 3

-dr and

= ^SR

3

Then

.

r

,

M

r

where

The

3

w2 =

differential

=

is

2

part (a)

we have

d r — m —-^ at 4

dA — = k(M — dA — = ki(M - A) A).

dt

22.

The

differential equation is

dt

-^ and

3

k-^j-

equation

r

Mm/R _

2

d r . From rr = ma = m-pr and at*

21.

T

it

Mm_ (b)

M

3

- y2

k^A.

—k

mM mM _ r R Aa