Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.8

Solve the following:

Question 1: dy/dx = (x + y + 1)²

Solution:

We have,

dy/dx = (x + y + 1)²

Putting x + y + 1 = v

Therefore, dv/dx - 1 = v²

⇒ dv/dx = v² + 1

⇒ 1/(v² + 1) dv = dx

Integrating both sides, we get

∫ 1/(v² + 1) dv = ∫ dx

Question 2: dy/dx cos (x - y) = 1

Solution: 

We have,

dy/dx cos (x - y) = 1

⇒ dy/dx = 1/cos(x - y)

Putting x - y = v

⇒ 1 - dy/dx = dv/dx

⇒ dy/dx = 1 - dv/dx

Therefore, 1 - dv/dx = 1/cos v

⇒ dv / dx = 1 - 1/cos v

⇒ dv/dx = (cos v - 1)/cos v

⇒ cos v/ (cos v - 1) dv = dx

Integrating both sides, we get

∫cos v/(cos v - 1) dv = ∫dx

⇒ -∫(cos v (1 + cosv)) / (1 - cos² v) dv =∫ dx

⇒ -∫(cos v (1 + cos v)) / (sin² v) dv = ∫ dx

⇒ -∫(cot v cosec v + cot² v) dv = ∫ dx

⇒ -∫ (cot v cosec v + cosec² v - 1) dv = ∫ dx

⇒ -(-cosec v - cot v - v)= x + C

⇒ cosec ( x - y ) + cot ( x - y ) + x - y = x + C

⇒ cosec ( x - y ) + cot ( x - y ) - y = C

⇒ ((1+cos ( x - y )) / sin ( x - y )) - y = C

⇒ cot (( x - y )/ 2) = y + C

Question 3: dy/dx = ((x - y) + 3)/ (2(x - y) + 5) 

Solution: 

We have,

dy/dx = ((x - y) + 3)/ (2(x - y) + 5) 

Putting x - y = v

⇒ 1 - dy/dx = dv/dx

⇒ dy/dx = 1 - dv/dx

Therefore, 1 - dv/dx = (v + 3)/ (2v + 5)

⇒ dv/dx = 1 - (v + 3)/ (2v + 5)

⇒ dv/dx = (2v + 5 - v - 3)/ 2v + 5

⇒ dv/dx = (v + 2) / (2v + 5)

⇒ (2v + 5)/(v + 2)dv = dx

Integrating both sides, we get

∫(2v + 5)/(v + 2) dv = ∫dx

 ⇒ ∫(2v + 4 + 1)/(v + 2) dv = ∫dx

⇒ ∫((2v + 4)/(v + 2) + 1/(v + 2))dv = ∫dx

⇒ 2∫dv + ∫1/(v + 2)dv = ∫dx

⇒ 2v + log |v + 2| = x + C

⇒ 2(x - y) + log | x - y + 2 | = x + C

Question 4: dy/dx = (x + y)²

Solution: 

We have,

dy/dx = (x + y)²

Let x + y = v

⇒ 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx - 1

Therefore, dv/dx - 1 = v²

⇒ dv/dx = v² + 1

⇒ 1/(v² + 1) dv = dx

Integrating both sides, we get

∫1/(v² + 1) dv = ∫dx

⇒ tan^-1 v = x + C

⇒ v = tan (x + C)

⇒ x + y = tan (x + C)

Question 5: (x + y)² dy/dx = 1

Solution: 

We have,

(x + y)² dy/dx = 1

⇒ dy/dx = 1/( x + y)²

Let x + y = v

⇒ 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx - 1

Therefore, dv/dx - 1 = 1/v²

⇒ dv/dx = 1/v² + 1

⇒ v²/(v² + 1) dv = dx

Integrating both sides, we get

∫v²/(v² + 1) dv = ∫dx

⇒ ∫v² + 1 - 1/(v² + 1) dv = ∫dx

⇒ ∫(1- 1/(v² + 1) dv = ∫dx

⇒ v - tan^-1 v = x + C

⇒ x + y - tan^-1 (x + y) = x + C

⇒ y - tan^-1 (x + y) = C

Question 6: cos² ( x - 2y) = 1 - 2dy/dx

Solution:  

We have,

cos² ( x - 2y ) = 1 - 2dy/dx 

⇒ 2dy/dx = 1 - cos² (x - 2y)

Let x - 2y = v

⇒ 1 - 2 dy/dx = dv/dx

⇒ 2 dy/dx = 1 - dv/dx

Therefore, 1 - dv/dx = 1 - cos² v

⇒ dv/dx = cos² v

⇒ sec² v dv = dx

Integrating both sides, we get

∫ sec² v dv = ∫dx

⇒ tan v = x - C

⇒ tan (x - 2y) = x - C

⇒ x = tan (x - 2y) + C

Question 7: dy/dx = sec(x + y)

Solution:  

We have,

dy/dx = sec(x + y)

⇒ dy/dx = 1/cos ( x + y)

Let x + y = v

⇒ 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx -1

Therefore, dv/dx - 1 = 1/cos v

⇒ dv/dx = (cos v + 1)/ cos v

⇒ cos v/(cos v + 1) dv = dx

Integrating both sides, we get

∫ cos v/(cos v + 1) dv = ∫ dx

⇒ ∫ cos v (1 - cos v)/(1 - cos² v ) dv = ∫ dx

⇒ ∫ cos v (1 - cos v)/sin² v  dv = ∫ dx

⇒ ∫ (cos v - cos^2 v)/sin² v  dv = ∫ dx

⇒ ∫(cot v cosec v - cot² v) dv = ∫ dx

⇒ ∫(cot v cosec v - cosec² v + 1) dv = ∫ dx

⇒ - cosec v + cot v + v = x + C

⇒ - cosec (x + y) + cot (x + y) + x + y = x + C

⇒ - cosec (x + y) + cot (x + y) + y = C

⇒ ((-1 + cos (x + y)) / sin (x + y)) + y = C

⇒ - tan ((x + y)/2) + y = C

⇒ y = tan((x + y)/2) + C

Question 8: dy/dx = tan (x + y)

Solution: 

We have,

dy/dx = tan (x + y)

dy/dx = sin (x + y)/cos (x + y)

Let x + y =v

Therefore, 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx - 1

Since, dv/dx -1 = sin v/cos v

⇒ dy/dx = sin v/cos v + 1

⇒ dy/dx = (sin v + cos v)/ cos v

⇒ cos v/(sin v + cos v) dv = dx

Integrating both sides, we get

⇒ ∫ cos v/(sin v + cos v) dv =∫ dx

⇒ 1/2 ∫ {(sin v + cos v) +  (cos v - sin v)}/(sin v + cos v) dv = ∫dx

⇒ 1/2 ∫ dv + 1/2 ∫ (cos v - sin v) / (sin v + cos v) dv = ∫dx

⇒ 1/2 v + 1/2 ∫ (cos v - sin v)/(sin v + cos v) dv = x

Putting sin v + cos v = t

⇒ (cos v - sin v) dv = dt

Therefore, 1/2 v + 1/2 ∫ dt/t = x

⇒ 1/2 v + 1/2 log |t| = x + C

⇒ (x + y) + 1/2 log |sin (x + y) + cos (x + y)| = x + C

⇒ 1/2 (y - x) + 1/2 log |sin (x + y) + cos (x + y)| = C

⇒ (y - x) + log |sin (x + y) + cos (x + y)| = 2C

⇒ y - x + log |sin (x + y) + cos (x + y)| = K                         (where K = 2C)

Question 9: (x + y) (dx - dy) = dx + dy

Solution: 

We have, 

(x + y) (dx - dy) = dx + dy

⇒ x dx + y dx -x dy - y dy = dx + dy

⇒ (x + y -1)dx = ( x + y +1) dy

⇒ dy/dx = (x + y -1)/(x + y + 1)

Let x + y = v

Therefore, 1 + dy/dx =dv/dx

⇒ dy/dx = dv/dx - 1

Therefore,  dv/dx -1 = (v - 1)/(v +1)

⇒ dv/dx = (v - 1)/(v +1) + 1

⇒ dv/dx = (v - 1 + v +1)/(v +1)

⇒ dv/dx = 2v / (v +1)

⇒ (v +1) / 2v dv = dx

Integrating both sides, we get

∫(v + 1)/2v dv = ∫ dx

⇒ 1/2 ∫dv + 1/2 ∫1/v dv = ∫dx

⇒ 1/2 v + 1/2 log |v| = x + C

⇒ 1/2 (x + y) + 1/2 log |x + y| = x + C

⇒ 1/2 (y - x) + 1/2 log |x + y| = C

Question 10: (x + y + 1)dy/dx = 1

Solution: 

We have,

(x + y + 1)dy/dx =1

⇒ dy/dx = 1/(x + y + 1)

Let x + y + 1 = v

Therefore, 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx - 1

Therefore, dv/dx - 1= 1/v

⇒ dv/dx = 1/v + 1

⇒ v/(v + 1) dv = dx

Integrating both sides, we get

∫ v/(v + 1) dv = ∫ dx

⇒ ∫ (v + 1 - 1)/(v + 1) dv = ∫ dx

⇒ ∫ (1 - 1/(v + 1))dv = ∫ dx

⇒ v - log |v + 1| = x + K

⇒ x + y + 1 - log |x + y+ 1 + 1| = x + K

⇒ y - log |x + y + 2| = K - 1

⇒ y - log |x + y + 2| = C₁        ( C₁ = K - 1)

⇒ y - C₁ = log |x + y + 2|

⇒ e^{y - C}₁ = x + y + 2

⇒ e^y / e^C₁ = x + y + 2

⇒ e^{- C}₁  e^y = x + y + 2

⇒ C e^y  = x + y + 2                  (C = e^{- C}₁)

⇒ x =  C e^y - y - 2  

Question 11: dy/dx + 1 = e^{x + y}  

Solution: 

We have,

dy/dx + 1 = e^{x + y}                      . . . (1)

Let x + y = t

⇒ 1 + dy/dx = dt/dx

Substituting the value of x + y = t and 1 + dy/dx = dt/dx from (1), we get

dt/dx = e^t 

⇒ e^{- t} dt = dx

⇒ - e^{- t} = x + C 

⇒ - e^{- (x + y)} = x + C         [Since t = x + y]

Summary

Exercise 22.8 in Chapter 22 of RD Sharma's Class 12 Solutions focuses on solving first-order linear differential equations. The practice problems cover a range of equations, including those with trigonometric functions, exponentials, and varying coefficients. Students are asked to find general solutions, particular solutions given initial conditions, and solve initial value problems. The equations generally follow the form dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x. The primary method used to solve these equations is the integrating factor method. The exercise helps students develop skills in recognizing the type of differential equation, applying the appropriate solving technique, and interpreting the results. It also reinforces understanding of concepts such as general vs. particular solutions, initial value problems, and the role of integrating factors in simplifying these equations.