MathCity.org Exercise 3.8 (Solutions)

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Exercise 3.8 (Solutions)Page177

Calculus and Analytic Geometry, MATHEMATICS 12 Available online @ , Version: 3.0

Question # 1

Check each of the following equations written

against the differential equation is its solution.

(i) x dy = 1+ y , y = cx -1

dx

(ii) x2 (2 y +1) dy -1 = 0 , y2 + y = c - 1

dx

x

(iii) y dy - e2x = 1, y2 = 2x + e2x + c dx

(iv)

1 dy x dx

-2y

=

0,

y = cex2

( ) (v)

dy dx

=

y2 + e- x

1

,

y = Tan ex + c

Solution

(i) x dy = 1+ y dx

x dy = (1+ y) dx dy = dx

1+ y x

Integrating both sides

dy 1+ y

=

dx x

ln (1+ y) = ln x + ln c

= ln cx 1+ y = cx

y = cx -1 Proved

(ii) x2 (2 y +1) dy -1 = 0

dx

x2 (2 y +1) dy = 1 x2 (2 y +1) dy = dx

dx

(2y

+1) dy

=

1 x2

dx

On integrating

(2y

+ 1)

dy

=

1 x2

dx

2 ydy + dy = x-2 dx

2 y2 + y = x-2+1 + c

2

-2 +1

y2 + y = x-1 + c - 1

y2 + y = c - 1 x

Proved

(iii) y dy - e2x = 1 dx

( ) y dy = 1+ e2x ydy = 1+ e2x dx dx On integrating

( ) ydy = 1+ e2x dx

y2 = x + e2x + c

2

22

y2 = 2x + e2x + c

y2 = 2x + e2x + c

(iv)

1 dy x dx

-2y

=

0

1 dy = 2 y dy = 2xy

x dx

dx

dy = 2xdx y

On integrating

dy y

=

2 xdx

ln y = 2 x2 + ln c 2

= x2 + ln c

= x2 ln e + ln c ln e = 1

= ln ex2 + ln c

ln y = ln cex2

y = cex2 Proved

(v)

dy = y2 +1

dx

e- x

dy y2 +1

=

exdx

Integrating both sides

dy y2 +1

=

exdx

Tan-1y = ex + c

( ) y = Tan ex + c

Solve the following differential equations:

Question # 2

dy dx

=

- y

Solution

dy dx

=

- y

dy y

=

- dx

On integrating

dy y

=

- dx

FSc-II / Ex- 3.8 - 2

ln y = - x + ln c

= - x ln e + ln c

ln e = 1

= ln e-x + ln c

ln y = ln ce-x y = ce-x

Question # 3 ydx + xdy = 0

Solution ydx + xdy = 0 ydx = - xdy

dx = - dy

x

y

On integrating ln x = - ln y + ln c

ln x

=

ln

c y

x= c y

xy = c

Question # 4

dy = 1- x dx y

Solution

Do yourself

Question # 5

dy dx

=

y x2

,

(

y

>

0)

Solution

dy dx

=

y x2

Integrating

dy = x-2dx y

dy y

=

x-2dx

ln y = x-2+1 + ln c -2 +1

ln y = x-1 + ln c - 1

ln y

=

- 1 + ln c x

ln y = - 1 ln e + ln c x

-1

= ln e x + ln c

ln y

=

-1

ln ce x

y

=

-1

ce x

Question # 6

sin

y

cosec

x

dy dx

= 1

Solution

sin y cosec x dy = 1 dx

sin y dy = dx cosec x

sin y dy = sin x dx

Integrating

sin y dy = sin x dx

- cos y = - cos x - c cos y = cos x + c

Question # 7

xdy + y ( x -1) dx = 0

Solution

xdy + y ( x -1) dx = 0 xdy = - y ( x -1) dx

dy = - x -1 dx

y

x

dy y

=

-

x x

-

1 x

dx

dy y

=

-

1 -

1 x

dx

On integrating

dy y

=

-

1

-

1 x

dx

ln y = - x + ln x + ln c

= - x ln e + ln x + ln c

= ln e-x + ln x + ln c

ln y = ln cxe-x y = cxe-x

Question # 8

x2 +1 = x dy ,( x, y > 0)

y +1 y dx

Solution

x2 +1 = x dy y +1 y dx

x2 +1 dx = y +1 dy

x

y

On integrating

x2 +1 x

dx

=

y

+1 y

dy

x2 x

+

1 x

dx

=

y y

+

1 y

dy



x

+

1 x

dx

=

1

+

1 y

dy

x

dx

+

1 x

dx

=

dy

+

1 y

dy

x2 + ln x = y + ln y - ln c 2

x2 ln e + ln x + ln c = y ln e + ln y 2

x2

ln e 2 + ln x + ln c = ln e y + ln y

x2

ln cxe 2 = ln ye y

x2

x2

cxe 2 = yey i.e. ye y = cxe 2

Question # 9

( ) 1 dy = 1 1+ y2

x dx 2

Solution

Do yourself

Question # 10

2x2 y dy = x2 -1 dx

Solution

Do yourself

Question # 11

dy + 2xy = x dx 2 y +1

Solution

dy + 2xy dx 2 y +1

=

x

dy = x - 2xy

dx

2y +1

=

x

1 -

2

2 y

y +

1

=

x

2

y +1-2 2y +1

y

dy dx

=

x

2

1 y+

1

(2 y +1) dy

=

x dx

Now do yourself

Question # 12

( ) x2 - yx2 dy + y2 + xy2 = 0 dx Solution

( ) x2 - yx2 dy + y2 + xy2 = 0 dx

FSc-II / Ex- 3.8 - 3

( ) x2 - yx2 dy = - y2 - xy2 dx

x2 (1- y) dy = - y2 (1+ x)

dx

1- y y2

dy

=

-

1+ x x2

dx

Now do yourself

Question # 13

sec2 x tan y dx + sec2 y tan x dy = 0

Solution

sec2 x tan y dx + sec2 y tan x dy = 0

sec2 x tan y dx = - sec2 y tan x dy

sec2 x dx = - sec2 y dy

tan x

tan y

On integrating

sec2 tan

x x

dx

=

-

sec2 tan

y y

dy

d dx

(

tan

tan x

x

)

dx

=

-

d dy

(

tan

y

)

dy

tan y

ln tan x = - ln tan y + ln c

ln tan x + ln tan y = ln c

ln (tan x tan y) = ln c

tan x tan y = c

Question # 14

y

-

x

dy dx

=

2

y2

+

dy dx

Solution

y

-

x

dy dx

=

2

y2

+

dy dx

y - x dy = 2 y2 + 2 dy

dx

dx

y - 2 y2 = 2 dy + x dy dx dx

y (1 - 2y)

=

(2

+

x

)

dy dx

dx 2+ x

=

dy

y (1 - 2y)

On integrating

dx 2+ x

=

dy

y (1 - 2y)

........... (i)

Now consider

1

y (1- 2 y)

=

A+ B y 1-2y



FSc-II / Ex- 3.8 - 4

1 = A(1- 2 y) + By .......... (ii)

Put y = 0 in (ii)

1 = A(1- 2(0)) + 0 A = 1

Put 1- 2 y = 0 2 y = 1 y = 1 in (ii) 2

1

=

0

+

B

1 2

B=2

So

1

y (1- 2 y)

=

1+ 2 y 1-2y

Using in (i)

dx 2+ x

=

1 y

+

2 1- 2

y

dy

=

1 y

dy

+

1

2 -2

y

dy

=

1 y

dy

-

1

- -

22 y dy

dx 2+ x

=

1 y

dy

-

d (1- 2

dx 1- 2y

y)

dy

ln (2 + x) = ln y - ln (1- 2 y) - ln c

ln (2 + x) + ln c = ln y - ln (1- 2 y)

ln c (2 + x)

=

ln

(1

y - 2

y

)

c(2+ x)

=

y

(1- 2 y)

y = c (2 + x)(1- 2 y)

Alternative ( )

dx 2+ x

=

1 y

+

2 1- 2

y

dx

=

1 y

dy

+

1

2 -2

y

dy

=

1 y

dy

-

2

2 y-

dy 1

dx 2+ x

=

1 y

dy

-

d dx

(

2

2y

y -1)

-1 dy

ln (2 + x) = ln y - ln (2 y -1) - ln c

ln (2 + x) + ln c = ln y - ln (2 y -1)

ln c (2 + x)

=

ln

(2

y

y -1)

c(2+ x)

=

y

(2 y -1)

i.e.

y

(2 y -1)

=

c(2+ x)

Review

tan x dx = ln sec x = - ln cos x cot x dx = ln sin x = - ln csc x sec x dx = ln sec x + tan x csc x dx = ln csc x - cot x

[

Question # 15

1

+

cos

x

tan

y

dy dx

=

0

Solution

1

+

cos

x

tan

y

dy dx

=

0

cos

x

tan

y

dy dx

=

- 1

tan y dy

=

-

1 cos

x

dx

tan y dy = - sec x dx

On integrating

tan y dy = - sec x dx

- ln cos y = - ln sec x + tan x - ln c

ln cos y = + ln sec x + tan x + ln c

ln cos y = ln c (sec x + tan x)

cos y = c (sec x + tan x)

Question # 16

y

-

x

dy dx

=

31

+

x

dy dx

Solution

y

-

x

dy dx

=

31

+

x

dy dx

y

-

x

dy dx

=

3

+

3x

dy dx

y-3

=

3x

dy dx

+

x

dy dx

=

(

3x

+

x

)

dy dx

y-3

=

4

x

dy dx

dx x

=

4

dy y-3

Now do yourself



Question # 17 sec x + tan y dy = 0 dx

Solution sec x + tan y dy = 0

dx tan y dy = - sec x

dx tan y dy = - sec x dx

Now do yourself as Question # 15

Question # 18

( ) ex + e-x dy = ex - e-x dx

Solution

( ) ex + e-x dy = ex - e-x dx

dy

=

ex ex

- +

e-x e-x

dx

On integrating

dy =

ex ex

- +

e- x e-x

dx

( ) y =

d ex + e-x

dx ex + e-x

dx

( ) y = ln ex + e-x + c

Question # 19 Find the general solution of the equation

dy - x = xy2 . Also find the perpendicular dx

solution if y = 1 when x = 0.

Solution

dy - x = xy2 dy = x + xy2

dx

dx

( ) dy = x 1+ y2 dx

dy 1+ y2

= x dx

1

dy +y

2

= x dx

Tan-1y = x2 + c 2

y

=

Tan

x2 2

+

c

FSc-II / Ex- 3.8 - 5

Question # 20

Solve the differential equation dx = 2x given dt

that x = 4 when t = 0

Solution

dx = 2x dx = 2dt

dt

x

dx x

= 2 dt

ln x = 2t + ln c

= ln e2t + ln c ln ex = x

ln x = ln ce2t

x = ce2t ...... (i)

When t = 0 then x = 4 , putting in (i) 4 = ce2(0) 4 = c e0

4 = c (1) c = 4

Putting in (i) x = 4e2t

Question # 21

Solve the differential equation ds + 2st = 0 . dt

Also find the perpendicular solution if s = 4e , when t = 0

Solution

ds + 2st = 0 dt

ds = - 2st ds = - 2t dt

dt

s

On integrating

ds s

=

- 2 t dt

ln s = - 2 t2 + ln c 2

= - t2 + ln c

= ln e-t2 + ln c ln ex = x

ln s = ln ce-t2

s = ce-t2 ....... (i)

When t = 0 then s = 4e , using in (i)

4e = ce-(0)2 4e = c (1) c = 4e

Putting in (i) s = 4e e-t2

s = 4e1-t2



FSc-II / Ex- 3.8 - 6

Question # 22 In a culture, bacteria increases at the rate proportional to the number of bacteria present. If

bacteria are 200 initially and are doubled in 2

hours, find the number of bacteria present four hours later. Solution

Number of bacteria initially = 200 No. of bacteria after two hours = 2(200)

= 400 No. of bacteria after four hours = 2(400)

= 800 Ans.

Question # 23

A ball is thrown vertically upward with a

velocity of 2450cm / sec. Neglecting air

resistance, find

(i) velocity of ball at any time t (ii) distance travelled in any time t

(iii) maximum height attained by the ball.

Solution i) When a body is projected upward its acceleration is -g . (where g = 980 cm/sec2)

i.e. acceleration =

dv dt

=

- g

,

where v is velocity of ball.

dv = - 980 dt

dv = - 980 dt

On integrating

dv = - 980 dt

v = - 980t + c1 ........ (i) Initially, when t = 0 then v = 2450 cm/sec

2450 = - 980 (0) + c1 c1 = 2450

Putting in (i)

v = - 980t + 2450

ii) Since velocity =

v

=

dx dt

where x is height of ball.

dx = - 980t + 2450 dt

dx = (-980t + 2450) dt

Integrating

dx = (-980t + 2450) dt

x

=

- 980

t2 2

+

2450t

+

c2

x = - 490 t2 + 2450t + c2 ....... (ii)

Initially, when t = 0 then x = 0

0 = - 490 (0) + 2450 (0) + c2 c2 = 0

Putting value of c2 in (ii)

x = - 490 t2 + 2450t + 0

x = 2450t - 490t2

iii) v = - 980t + 2450

When body is at max. height then v = 0 - 980t + 2450 = 0

980t = 2450 t = 2450 980

t = 2.5 sec

Since x = 2450t - 4980t2

When t = 2.5sec

x = 2450(2.5) - 490(2.5)2

= 6125 - 3062.5 = 3062.5 Hence ball attains max. height of 3062.5 cm.

Book: Exercise 3.8 ,page 177. Calculus and Analytic Geometry Mathematic 12 Punjab Textbook Board, Lahore.

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