RD Sharma Solutions for Class 12 Maths Exercise 22.6 Chapter 22 Differential Equations are available here. If students want to score higher in Maths, it requires the right amount of practice for every topic. More marks can be achieved by referring to RD Sharma Solutions for Class 12 Maths Chapter 22. Problems are solved step by step by BYJU’S experts with neat explanations.
RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 6
Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 6
EXERCISE 22.6
Question. 1
Solution:
From the question, it is given that,
(dy/dx) + ((1 + y2)/y) = 0
Transposing we get,
dy/dx = – (1 + y2)/y
By cross multiplication,
(y/(1 + y2)) dy = – dx
Integrating on both sides, we get,
∫(y/(1 + y2)) dy = ∫-dx
∫(2y/(1 + y2)) dy = -2 ∫dx
log (1 + y2) = – 2x + c1
Therefore, ½log [1 + y2] + x = c
Question. 2
Solution:
From the question, it is given that,
(dy/dx) = ((1 + y2)/y3)
By cross multiplication,
(y3/(1 + y2)) dy = dx
Integrating on both sides, we get,
∫(y – (y/(1 + y2)) dy = ∫dx
∫ydy – ∫(y/(1 + y2)) dy = ∫ dx
∫ydy – ½ ∫(2y/(1 + y2)) dy = ∫ dx
(y2/2) – ½ log [y2 + 1] = x + c
Question. 3
Solution:
From the question, it is given that,
(dy/dx) = sin2 y
By cross multiplication,
dy/sin2 y = dx
We know that, (1/sin x) = cosec x
cosec2 y dy = dx
Integrating on both sides, we get,
∫cosec2 y dy = ∫dx + c
– cot y = x + c
Question. 4
Solution:
From the question, it is given that,
(dy/dx) = (1 – cos 2y)/(1 + cos 2y)
We know that, 1 – cos 2y = 2 sin2y and 1 + cos 2y = 2 cos2 y
So, dy/dx = (2 sin2 y)/(2 cos2 y)
Also, we know that sin θ/cos θ = tan θ
By cross multiplication,
dy/tan2 y = dx
Integrating on both sides, we get,
∫cot2 y dy = ∫dx
∫ (cosec2 y – 1) dy = ∫dx
– cot y- y + c = x
c = x + y + cot y
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