Hello Students. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 9.2.
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NCERT Solutions for Maths Class 12 Exercise 9.1
NCERT Solutions for Maths Class 12 Exercise 9.3
NCERT Solutions for Maths Class 12 Exercise 9.4
NCERT Solutions for Maths Class 12 Exercise 9.5
NCERT Solutions for Maths Class 12 Exercise 9.2
In each of the
Exercises 1 to 10, verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation.
Maths Class
12 Ex 9.2 Question 1.
y = ex + 1 :
y’’ – y’ = 0
Solution:
The given
function is y = ex + 1 … (i)
Differentiating equation (i) w.r.t. x, we get
y’ = ex … (ii) and y’’ =
ex … (iii)
Subtracting equation (ii) from equation (iii), we get y’’ – y’ = 0
Hence, the function y = ex + 1 is the solution of y’’ – y’ = 0.
Maths Class
12 Ex 9.2 Question 2.
y = x2 + 2x + C : y’
– 2x – 2 = 0
Solution:
The given
function is y = x2 + 2x + C … (i)
Differentiating equation (i) w.r.t. x, we get
y’ = 2x + 2 ⇒ y’ – 2x – 2 = 0 … (ii)
Hence, the function y = x2 + 2x + C is the solution of y’
– 2x – 2 = 0.
Maths Class
12 Ex 9.2 Question 3.
y = cos x + C :
y’ + sin x = 0
Solution:
The given
function is y = cos x + C … (i)
Differentiating equation (i) w.r.t. x, we get
y’ = –sin x ⇒ y’ + sin x = 0 … (ii)
Hence, the function y = cos x + C is the solution of y’ + sin x = 0.
Maths Class
12 Ex 9.2 Question 4.
Solution:
The
given function is y = √(1 + x2) … (1)
Differentiating equation (i) w.r.t. x, we get
Hence, the function y = √(1 + x2) is the solution of y’ = xy/(1 + x2).
Maths Class
12 Ex 9.2 Question 5.
y = Ax
: xy’ = y (x
≠ 0)
Solution:
The given
function is y = Ax … (i)
Differentiating equation (i) w.r.t. x, we get
y’ = A … (ii)
Dividing equation (ii) by equation (i), we get
y’/y = 1/x ⇒ xy’ = y
Hence, the function y = Ax is the solution of xy’ = y (x ≠ 0).
Maths Class
12 Ex 9.2 Question 6.
Solution:
The given
function is y = x sin
x … (i)
Differentiating equation (i) w.r.t. x, we get
y’ = x cos
x + sin x … (ii)
Multiplying
equation (ii) by x, we get
xy’ = x2 cos x + x sin x
xy = log y
+ C :
Solution:
The given
function is xy = log y + C … (i)
Differentiating equation (i) w.r.t. x, we get
y – cos y = x :
(y sin y + cos y + x) y’
= y
Solution:
The given
function is y – cos y = x … (i)
Differentiating equation (i) w.r.t. x, we get
y’ – (–sin y)
y’ = 1 ⇒ y’(1 + sin y) = 1 … (ii)
Multiplying equation (ii) by y,
we get
y’y(1 + sin y) = y ⇒ y’(y + y
sin y) = y ⇒ y’(x
+ cos y + y sin y) = y
Hence, the function y – cos y = x
is the solution of y’(x + cos y + y sin y) = y.
Maths Class
12 Ex 9.2 Question 9.
x + y = tan-1 y
: y2 y’ + y2 +
1 = 0
Solution:
The given
function is x + y = tan-1 y … (i)
Differentiating equation (i) w.r.t. x, we get
1 + y’ = 1/(1 + y2) × y’ ⇒ (1 + y’) (1 + y2) = y’
⇒ 1 + y2 + y’ + y2 y’ = y’ ⇒ y2 y’ + y2 +
1 = 0
Hence, the function x + y = tan-1 y is the solution of y2 y’ + y2 + 1 = 0.
Maths Class
12 Ex 9.2 Question 10.
The number of
arbitrary constants in the general solution of a differential equation of
fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
Solution:
(D) The general
solution of a differential equation of fourth order has 4 arbitrary constants.
Because it contains the same number of arbitrary constants as the order of
differential equation.
Maths Class
12 Ex 9.2 Question 12.
The number of
arbitrary constants in the particular solution of a differential equation of
third order are:
(A) 3
(B) 2
(C) 1
(D) 0
Solution:
(D) Number of arbitrary
constants = 0
Because particular solution is free from arbitrary constants.
Related Links:
NCERT Solutions for Maths Class 12 Exercise 9.1
NCERT Solutions for Maths Class 12 Exercise 9.3