RD Sharma Solutions for Class 12 Maths Exercise 22.5 Chapter 22 Differential Equations are provided here. It is the best study material for those students who are having difficulties solving problems. These can help students clear doubts quickly and help in understanding the topic effectively. RD Sharma Solutions for Class 12 Maths Chapter 22 can help students to score well in the examination.
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Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 5
EXERCISE 22.5
Question. 1
Solution:
From the question, it is given that,
dy/dx = x2 + x – (1/x)
Integrating on both sides, we get,
∫dy = ∫(x2 + x – (1/x)) dx
We know that ∫ xn dx = x(n + 1)/(n + 1)
y = (x3/3) + (x2/2) – log x + c
Question. 2
Solution:
From the question, it is given that,
dy/dx = x5 + x2 – (2/x)
Integrating on both sides, we get,
∫dy = ∫(x5 + x2 – (2/x)) dx
We know that ∫ xn dx = x(n + 1)/(n + 1)
y = (x6/6) + (x3/3) – 2 log x + c
Question. 3
Solution:
From the question, it is given that,
dy/dx + 2x = e3x
Transposing we get,
dy/dx = e3x – 2x
Integrating on both sides, we get,
∫dy = ∫(e3x – 2x) dx
We know that ∫ xn dx = x(n + 1)/(n + 1)
y = (e3x/3) – (2x2/2) + c
y = (e3x/3) – x2 + c
Therefore, y + x2 = 1/3 (e3x) + c
Question. 4
Solution:
From the question, it is given that,
(x2 + 1)dy/dx = 1
By cross multiplication,
dy = dx/(x2 + 1)
Integrating on both sides, we get,
∫dy = ∫(dx/(x2 + 1)
We know that ∫ dx/(x2 + 1) = tan -1 x + c
Therefore, y = tan -1 x + c
Question. 5
Solution:
From the question, it is given that,
dy/dx = (1 – cos x)/(1 + cos x)
We know that (1 – cos x) = 2 sin2 (x/2) and (1 + cos x) = 2 cos2 (x/2)
So,
dy/dx = (2 sin2 (x/2))/(2 cos2 (x/2))
Also, we know that = sin θ/cos θ = tan θ
dy/dx = tan2 (x/2)
By cross multiplication,
dy = tan2 (x/2) dx
Integrating on both sides, we get,
∫dy = ∫tan2 (x/2) dx
We know that, sec2 x – 1 = tan2 x
∫dy = ∫sec2 (x/2) – 1 dx
y = 2 tan (x/2) – x + c
Therefore, y = 2 tan (x/2) – x + c
Question. 6
Solution:
From the question, it is given that,
(x + 2)dy/dx = x2 + 3x + 7
By cross multiplication,
dy = [(x2 + 3x + 7)/(x + 2)] dx
On dividing, we get,
dy = [x + 1 + (5/(x +2))] dx
Integrating on both sides, we get,
∫dy = ∫[x + 1 + (5/(x +2))] dx
Therefore, y = (x2/2) + x + 5 log (x + 2) + c
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