NCERT Solutions for Class 12 Maths Exercise Miscellaneous Chapter 9 Differential Equations – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 Exercise Miscellaneous (Ex Misc) 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 Miscellaneous 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 Misc.) Exercise Miscellaneous
1. For each of the differential equations given below, indicate its order and degree (if defined):
(i) 
(ii) 
(iii) 
The highest order derivative present in this differential equation is 
and hence order of this differential equation if 2.
The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative is 1.
Therefore, Order = 2, Degree = 1
(ii) Given: Differential equation
The highest order derivative present in this differential equation is 
and hence order of this differential equation if 1.
The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative is 3.
Therefore, Order = 1, Degree = 3
(iii) Given: Differential equation
The highest order derivative present in this differential equation is 
and hence order of this differential equation if 4.
The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.
Therefore, Order = 4, Degree not defined
2. For each of the exercises given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:
(i) 
(ii) 
(iii) 
(iv) 
……….(i) To verify: Function (i) is a solution of D.E. 
……….(ii)
Differentiating both sides of eq. (i) w.r.t. 
Again differentiating both sides w.r.t.
Putting 
from eq. (i), we have,
Therefore, Function given by eq. (i) is a solution of D.E. (ii).
(ii) The given function is 
……….(i)
To verify: Function given by (i) is a solution of D.E. 
…….(ii)
From (i), 
[By eq. (i)] ……….(iii)
[Using eq. (iii) and (i)]
Therefore, Function given by eq. (i) is a solution of D.E. (ii).
(iii) The given function is 
………(i)
To verify: Function given by eq. (i) is a solution of D.E. 
…(ii)
From eq. (i), 
[Using eq. (i)]
Therefore, Function given by eq. (i) is a solution of D.E. (ii).
(iv) The given function is 
……….(i)
To verify: Function given by eq. (i) is a solution of D.E. 
……(ii)
Differentiating both sides of eq. (i) w.r.t.
Putting 
from eq. (i), we get
Therefore, Function given by eq. (i) is a solution of D.E. (ii).
3. Form the differential equation representing the family of curves 
where ia an arbitrary constant.
where
……….(i)
Here number of arbitrary constants is one only 
So, we will differentiate both sides of equation only once, w.r.t.
……….(ii)
Dividing eq. (i) by eq. (ii), we have
=> 
4. Prove that is the general equation of the differential equation where is a parameter.
is the general equation of the differential equation 
where 
is a parameter.
……….(i) Here each coefficient of 
and 
is of same degree, i.e., 3, therefore differential equation looks to be homogeneous.
……….(ii)
Therefore, the given differential equation is homogeneous.
Putting 
Putting these values in eq. (ii),
=> 
[Separating variables]
Integrating both sides
, 
……….(iii)
Now forming partial fraction of 
= 
= 
= 
……….(iv)
Comparing coefficients of like powers of
A – B – C = 1 ……….(v)
A + B – D = 0 ……….(vi)
A – B + C = –3 ……….(vii)
Constants A + B + D = 0 ………(viii)
Now eq. (v) – eq. (vii)
– 2C = 4
C = – 2
Eq. (vi) – eq. (viii)
– 2D = 0 D = 0
Putting C = – 2 in eq. (v), A – B + 2 = 1
A – B = –1 ……….(ix)
Putting D = 0 in eq. (vi) A + B = 0 ……….(x)
Adding eq. (ix) and (x) 2A = –1
A = 
From eq. (x), B = –A = 
Putting the values of A, B, C and D in eq. (iv), we have
= 
Putting this value in eq. (iii),
=> 
Squaring both sides and cross-multiplying,
=> 
Putting 
=> 
where 
5. For the differential equation of the family of the circles in the first quadrant which touch the coordinate axes.
where 
is the radius of the circle. Equation of the circle is
……….(i)
Differentiating with respect to
=> 
Substituting value of in eq. (i),
(x−y)2(1+(y′)2)=(x+yy′)2(x−y)2(1+(y′)2)=(x+yy′)2
6. Find the general solution of the differential equation
Integrating both sides
sin-1y = -sin-1x + c
7. Show that the general solution of the differential equation is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.
Integrating both sides,
……….(i)
Now 
[Completing the squares]
Therefore, 
= 
Similarly, 
Putting these values in eq. (i),
[Multiplying by 
]
where 
dividing by 
where A = 
8. Find the equation of the curve passing through the point whose differential equation is
whose differential equation is 
Integrating both sides,
……….(i)
Now, curve (i) passes through 
Therefore, putting 
in eq. (i),
Putting in eq. (i),
9. Find the particular solution of the differential equation given that when
when 
Dividing every term by 
we have
Integrating both sides,
……….(i)
Now to evaluate 
,
putting 
= 
Putting this value in eq. (i), 
……….(ii)
Now putting 
in eq. (ii),
Putting in eq. (ii),
10. Solve the differential equation:
……….(i)
It is not a homogeneous differential equation because of presence of only 
as a factor,
yet it can be solved by putting 
i.e., 
Putting these values in eq. (i), we get
evdv=dyevdv=dy
Integrate both sides, we get
11. Find the particular solution of the differential equation given that when
when
……….(i)
Putting 
Putting this value in eq. (i),
= 
Integrating both sides,
∫(t+1t)dt=2∫1.dx∫(t+1t)dt=2∫1.dx
Putting ,
……….(ii)
Now putting 
in eq. (ii),
Putting in eq. (ii),
12. Solve the differential equation:
Comparing this equation with 
,
P = 
and Q = 
I.F. = 
The general solution is
y (I.F.) = 
ye2x√=∫e−2x√x√e2x√dx+cye2x=∫e−2xxe2xdx+c
ye2x√=∫1x√dx+cye2x=∫1xdx+c
ye2x√=2x−−√+cye2x=2x+c
13. Find the particular solution of the differential equation given that when
given that 
when 
Comparing this equation with ,
P = 
and Q = 
I.F. = 
The general solution is
y (I.F.) =
……….(i)
Now putting 
in eq. (i),
Putting in eq. (i),
14. Find the particular solution of the differential equation given that when
Integrating both sides,
Putting 
Putting ,
where C = 
……….(i)
Putting 
in eq. (i),
= C
C = 1
Putting C = 1 in eq. (i),
This solution may be written as
= 
where expresses as an explicit function of 
15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?
According to the question, Rate of increase of population of the village is proportional to the number of inhabitants.
where 
> 0 because of increase and is the constant of proportionality
[Separating variables]
Integrating both sides, 
……….(i)
Now Population of the village was P = 20,000 in the year 1999.
Let us take the base year 1999 as 
Putting 
and P = 20000 in eq. (i), 
Now putting in eq. (i), 
………..(ii)
Again Population of the village was P = 25,000 in the year 2004, when 
Putting 
and P = 25000 in eq. (ii), 
Putting value of in eq. (ii), 
………..(iii)
To find the population in the year 2009, 
Putting 
in eq. (iii), 
=>P=2516×20000=>P=2516×20000
25 x 1250 = 31250
Choose the correct answer:
16. The general solution of the differential equation 
is:
(A) 
= C
(B) 
(C) 
(D) 
[Separating variables]
Integrating both sides, 
where C = 
Therefore, option (C) is correct.
17. The general equation of a differential equation of the type is:
is:
(A) 
(B) 
(C) 
(D) 
is 
where 
= 
Therefore, option (C) is correct.
18. The general solution of the differential equation is:
(A) 
(B) 
(C) 
(D)
Comparing with 
P = 1 and Q = 
I.F. = 
Solution is 
Therefore, option (C) is correct.