Mathematics_Today_-_October_2016

⇒−^1 +=

3 v^3

log |vx| c

⇒−x +=

y

yc

3

3 3 log | |

This is the required solution of the given differential

equation.

17. Let P be the principal at any time t. Then,
    dP
    dt

== 5 P^5 P

100

%of

⇒ dP=

dt

P

20

⇒^11 =

P 20

dP dt

Integrating both sides, we get

11

P 20

∫ dP=∫ dt

⇒ logPt C=+^1 log

(^20)
⇒ logP=
C
(^1) t
20
⇒ P = Cet/20 ...(1)
When t = 0, we have P = 1000
Putting these values in (1), we get 1000 = C
⇒ P = 1000 et/20 ...(2)
Let T years be the time required to double the
principal i.e., at t = T, P = 2000. Substituting these
values in (2), we get
2000 = 1000eT/20
⇒ eT/^20 = 2 ⇒ T =logee⇒ T= log
20
2202
Hence, the principal doubles in 20 loge 2 years.

18. We h a v e y dx –(x + 2y^2 )dy = 0

or dx or

dy

xy

y

dx

dy

x

y

= + + ⎛− y

⎝⎜

⎞

⎠⎟

(^21) = 2
2
...(1)
This differential equation is of the form
dx
dy
+=Px Q, where
P
y
=−^1 andQy= 2
Now, I.F.== = =∫ ∫
− −
ee e
y
Pdy ydy y
1
log^1
∴ Solution is given by x
y
Q
y
⋅^11 =∫ ⋅ dy C+
⇒ x=∫ ⋅ += +
y
y
y
2 1 dy C 2 y C^ ⇒^ x = 2y^2 + Cy^
This is the general solution of the given differential
equation.

19. We h a v e
    220 xy y^22 x dy
    dx

+ − =

...(1)

and y(1) = 2

From (1), dy

dx

xy y

x

=^2 +

2

2

2 ...(2)

This is a linear homogeneous differential equation.

Put y = vx, then dy

dx

vxdv

dx

=+

∴ Equation (2) becomes

vxdv

dx

+ =^2 + =+

2

1

2

vv^22

vv

⇒ xdv=

dx

(^1) v
2
(^2) ⇒− 20 +=
2
dv
v
dx
x
Integrating, we get
2
v
+=log | |xc
⇒^2 x+=
y
log | |xc ...(3)
Since y(1) = 2
∴ From (3),^21
2
× +=log 1 c ⇒ c = 1
⇒ 2 x 1
y
+=log | |x ⇒ 2 x – y + ylog|x| = 0
This is the required particular solution.

20. We h a v e dy
    dx

+= +yxx xxcot^2 cot 2 ...(1)

This is a linear D.E. of the form dy

dx

+=Py Q

Where P = cot x and Q = x^2 cotx + 2x

∴ I.F.==ee∫∫Pdx cotx dx

=elog sinx = sinx

∴ The solution of (1) is given by

yxx xxxdxc⋅sin =+ +∫(^2 cot 2 ) sin

=++∫x xdx ∫x xdx c

(^2) cos 2 sin
=x x−⋅∫ x xdx++∫x xdx c
(^2) sin 22 sin sin
= x^2 sin x + c
⇒ y = x^2 + c · cosec x ...(2)
Since,,yx== 0 when
2
π
∴ From (2), 0
44
22
=+ππcc⇒ =−
⇒ yx=^2 − x
2
4
π cosec
This is the required particular solution of the given
differential equation.
””
MPP-4 CLASS XI ANSWER KEY

1. (d) 2. (b) 3. (c) 4. (a) 5. (c)


2. (b) 7. (a, b) 8. (c) 9. (b,c,d) 10. (a,b,c)


3. (a,b) 12. (b,c) 13. (a,b,c) 14. (c) 15. (d)


4. (d) 17. (7) 18. (6) 19. (6) 20. (3)