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.
Available online at in PDF Format (Picture format to view online). Updated: 11-9-2017.
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