NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-2/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2

Ex 9.2 Class 12 NCERT Solutions Question 1.
y = e^x + 1 : y”- y’ = 0
Solution:
y = e^x + 1
Differentiating w.r.t. x, we get
y’ = e^x
i.e., y” = y’
⇒ y” – y’ = 0
Hence y = e^x is a solution of y” – y’ = 0.

Ex 9.2 Class 12 Maths Ncert Solutions Question 2.
y = x² + 2x + C : y’ – 2x – 2 = 0
Solution:
y = x² + 2x + C
Differentiating w.r.t. x, we get y’ = 2x + 2
i.e., y’ – 2x – 2 = 0
Hence y = x² + 2x + C is a solution of y’ – 2x – 2 = 0

Question 3.
y = cos x + C : y’ + sin x = 0
Solution:
y = cos x + C
Differentiating w.r.t., x, we get y’ = – sin x
i.e., y’ + sin x = 0
Hence y = cos x + C is a solution of y’ + sin x = 0

Question 4.
y = \(\sqrt{1+x^{2}} \quad: \quad y^{\prime}=\frac{x y}{1+x^{2}}\)
Solution:
\(\sqrt{1+x^{2}}\)
Differentiating w.r.t. x, we get

Question 5.
y – Ax : xy’ = y (x ≠ 0)
Solution:
y = Ax … (1)
Differentiating w.r.t., x, we get y’ = A
i.e., y’ = \(\frac { y }{ x }\) [From(1), A = \(\frac { y }{ x }\)]
i.e., xy’ = y (x ≠ 0)
∴ y = Ax is a solution of xy’ = y.

Question 6.
y = x sin x : xy’= y + x \(\sqrt{x^{2}-y^{2}}\)
(x ≠ 0 and x > y or x < – y)
Solution:
y = x sinx … (1)
Differentiating (1) w.r.t. x, we get
y’ = x cos x + sin x … (2)
From (1) sin x = \(\frac { y }{ x }\)

∴ y = x sinx is a solution of xy’ = x\(\sqrt{x^{2}-y^{2}}\) + y

Question 7.
Ay = logy + C : y’ = \(\frac{y^{2}}{1-x y}(x y \neq 1)\)
Solution:
xy = logy + C
Differentiating w.r.t. x, we get
xy’ + y = \(\frac { 1 }{ y }\) y’
i.e., xyy’ + y² = y’
y’ – xyy = y²
y'(1 – xy) = y²
i.e., y’ = \(\frac{y^{2}}{1-x y}\)
Hence xy = logy + C is a solution of y’ = \(\frac{y^{2}}{1-x y}\)

Question 8.
y – cos y = x : (ysiny + cosy + x)y’ = y.
Solution:
y – cosy = x … (1)
Differentiating (1) w.r.t. x, we get
y’ + sin y . y’ = 1
i. e., yy’ + yy’sin y = y
(x + cos y)y’ + yy’sin y = y
[since from (1) y = x + cosy]
xy’ + y’cosy+ yy’siny = y
(y sin y + cos y+ x)y’ = y
Hence y – cosy = x is a solution of
(y sin y + cos y + x)y’ = y

Question 9.
x + y = tan^-1 y : y²y’ + y² + 1 = 0
Solution:
x + y = tan^-1
Differentiating w.r.t. x, we get,
1 + y’ = \(\frac{1}{1+y^{2}}\)
i.e., (1 + y²)(1 + y’) = y ‘
1 + y’ + y² + y²y’ = y’
i.e., y²y’ + y² + 1 = 0
Hence x + y = tan^-1 is a solution of y²y’ + y² + 1 = 0

Question 10.
\(y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)\)
Solution:
y = \(\sqrt{a^{2}-x^{2}}\) , x ∈ (- a, a)
Differentiating w.r.t. x, we get

Hence y = \(\sqrt{a^{2}-x^{2}}\) is a solution of x + y\(\frac { dy }{ dx }\) = 0

Question 11.
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. 4
The general solution of a differential equation contains as many arbitrary constants as the order of the differential equation. Since the differential equation is of order 4, its solution contains 4 arbitrary constants.

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. 0
The particular solution is free from the arbitrary constants.