NCERT Solutions for Class 12 Maths Exercise 9.4 Chapter 9 Differential Equations – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4 (Ex 9.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 9 Differential Equations Exercise 9.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks.
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.4) Exercise 9.4
For each of the differential equations in Questions 1 to 4, find the general solution:
1. 
Integrating both sides,
2. 
Integrating both sides,
=> 
3.
Integrating both sides,
=> 
, where 
4.
Dividing by 
we have
5.
Integrating both sides,
=> 
6.
[Separating variables]
Integrating both sides,
=> 
7.
[Separating variables]
Integrating both sides,
=> 
Putting 
on L.H.S., we get
Now eq. (i) becomes
[If all the terms in the solution of a differential equation involve log, it is better to use 
or
instead of 
in the solution.]
where 
8.
[Separating variables]
Integrating both sides
=> 
y−4−4=−x−4−4+cy−4−4=−x−4−4+c
x−4+y−4=−4cx−4+y−4=−4c
, where 
9.
Integrating both sides,
Applying product rule,
= 
……….(i)
To evaluate 
Putting, 
differentiate 
= 
Putting this value in eq. (i), the required general solution is
10.
Dividing each term by 
we get
[Separating variables]
Integrating both sides,
For each of the differential equations in Question 11 to 14, find a particular solution satisfying the given condition:
11. 
when 
[Separating variables]
Integrating both sides,
…(i)
Let 
[Partial fraction] ……….(ii)
Comparing the coefficients of 
on both sides, A + B = 2 ……….(iii)
Comparing the coefficients of 
on both sides, B + C = 1 ……….(iv)
Comparing constants on both sides, A + C = 0 ……….(v)
From eq. (iii) – (iv), we haveA – C = 1 ……….(vi)
Adding eq. (v) and (vi), we have 2A = 1
A = 
From eq. (v), we have C = – A = 
Putting the value of C in eq. (iv), B – = 1
B = 1 + = 
Putting the values of A, B, and C in eq. (ii), we have
= 
= 
Putting this value in eq. (i),
……….(vii)
Now, when 
, putting these values in eq. (vii),
Putting value of in eq. (vii), the required general solution is
12. when
when
Integrating both sides,
=> 
……….(i)
Let 
……….(ii)
Comparing the coefficients of on both sides, A + B + C = 0 ……….(iii)
Comparing the coefficients of on both sides, 
B + C = 0
C = B ……….(iv)
Comparing constants on both sides, A = 1
A = 1 ……….(v)
Putting A = 1 and C = B in eq. (iii),
1 + B + B = 0
2B = 1
B =
From eq. (iv),C = B =
Putting the values of A, B and C in eq. (ii), we get
Putting this value in eq. (i),
……….(v)
Now, putting 
when 
in eq. (v), we get
Putting the value of in eq. (v), the required general solution is
[NOTE: You can also do, to evaluate 
= 
= 
Put 
]
13. when
when
when 
dy = (cos-1 a)dx
Integrating both sides,
=> 
……….(i)
Now putting 
when in eq. (i), we get
Putting in eq. (i),
14. when
when
[Separating variables]
Integrating both sides,
=> 
where 
……….(i)
Now putting and in eq. (i), we get 
Putting C = 1 in eq. (i), we get the required general solution
15. Find the equation of the curve passing through the point (0, 0) and whose differential equation is
Integrating both sides
=> 
………(i)
where I = 
……….(ii)
Applying product rule,
Again applying product rule,
[By eq. (ii)]
2I = 
I = 
Putting this value of I in eq. (i), we get
……….(iii)
Now putting and 
in eq. (iii)
Putting the value of in eq. (iii), we get the required general solution
16. For the differential equation find the solution curve passing through the point
find the solution curve passing through the point 
[Separating both sides]
Integrating both sides
=> 
=>∫y+2−2y+2dy=∫(xx+2x)dx=>∫y+2−2y+2dy=∫(xx+2x)dx
=>∫y+2y+2−2y+2dy=∫(1+2x)dx=>∫y+2y+2−2y+2dy=∫(1+2x)dx
=>∫(1−2y+2)dy=∫(1+2x)dx=>∫(1−2y+2)dy=∫(1+2x)dx
……….(i)
Now putting 
in eq. (i),
Putting this value of in eq. (i) to get the required solution curve
17. Find the equation of the curve passing through the point given that at any point on the curve the product of the slope of its tangent and coordinate of the point is equal to the coordinate of the point.
coordinate of the point is equal to the 
coordinate of the point.
be any point on the required curve.According to the question, Slope of the tangent to the curve at P 
Integrating both sides,
=> 
where 
Now it is given that curve passes through the point 
.
Therefore, putting and 
in this equation, we get C = 4
Putting the value of C in the equation ,
18. At any point of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point Find the equation of the curve given that it passes through
Find the equation of the curve given that it passes through
of the required curve= 2. Slope of the line joining the point of contact P to the given point A
[Separating variables]
Integrating both side,
=> 
where 
……….(i)
Now it is given that curve (i) passes through the point 
Therefore, putting 
and in eq. (i),
=> 1+3 = C(-2+4)2
4 = 4C
C = 1
Putting C = 1 in eq. (i), we get the required solution,
19. The volume of the spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after seconds.
be the radius of the spherical balloon at time 
Given: Rate of change of volume of spherical balloon is constant = 
(say) 
[Separating variables]
Integrating both sides,
=> 
……….(i)
Now it is given that initially radius is 3 units, when 
Therefore, putting 
in eq. (i),
=> 4π×273=c4π×273=c
……….(ii)
Again when 
sec, then 
units
Therefore, putting and in eq. (i),
=> 
[From eq. (ii)]
……….(iii)
Putting the value of and in eq. (i), we get
20. In a bank principal increases continuously at the rate of per year. Find the value of if Rs 100 double itself in 10 years.
if Rs 100 double itself in 10 years. 
years.According to the given condition, rate of increase of principal per year = 
(of principal) 
[Separating variables]
Integrating both sides,
……….(i)
[Since P being principal > 0, hence 
]
Now initial principal = Rs 100 (given), i.e., when 
then P = 100
Therefore, putting P = 100 in eq. (i),
Putting in eq. (i),
……….(ii)
Now putting P = double of itself = 2 x 100 = Rs 200, when 
years (given)
21. In a bank principal increases at the rate of per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years?
years.According to the given condition, rate of increase of principal per year = 
(of principal) 
[Separating variables]
Integrating both sides,
……….(i)
[Since P being principal > 0, hence ]
Now initial principal = Rs 1000 (given), i.e., when then P = 1000
Therefore, putting P = 1000 in eq. (i),
Putting in eq. (i),
……….(ii)
Now putting years (given)
P = 1000 x 1.648 = Rs 1648
22. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.
be the bacteria present in the culture at time hours.According to the question,Rate of growth of bacteria is proportional to the number present
is proportional to
where 
is the constant of proportionality
Integrating both sides,
=> 
……….(i)
Now it is given that initially the bacteria count is 
(say) = 1,00,000
when then 
Putting these values in eq. (i)
Putting in eq. (i), we get
=> 
……….(ii)
Now it is given also that the number of bacteria increased by 10% in 2 hours.
Therefore, increase in bacteria in 2 hours = 
= 10,000
x, the amount of bacteria at 
= 1,00,000 + 10,000 = 1,10,000 = 
(say)
Putting 
and in eq. (ii),
Putting this value of in eq. (ii), we get,
[when 
]
hours
23. The general solution of the differential equation is:
(A) 
(B) 
(C) 
(D) 
[Separating variables]
Integrating both sides,
where 
which is required solution
Therefore, option (A) is correct.