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# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.8

• Last Updated : 03 Jan, 2021

### 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

â‡’ dv/dx = v2 + 1

â‡’ 1/(v2 + 1) dv = dx

Integrating both sides, we get

âˆ« 1/(v2 + 1) dv = âˆ« dx

### Question 2: dy/dx cos (x – y) =1

Solution:

We have,

dy/dx cos (x – y) = 1

â‡’ dy/dx = 1/cos(x – y)

Putting x – y = v

â‡’ 1 – dy/dx = dv/dx

â‡’ dy/dx = 1 – dv/dx

Therefore, 1 – dv/dx = 1/cos v

â‡’ dv / dx = 1 – 1/cos v

â‡’ dv/dx = (cos v – 1)/cos v

â‡’ cos v/ (cos v – 1) dv = dx

Integrating both sides, we get

âˆ«cos v/(cos v – 1) dv = âˆ«dx

â‡’ -âˆ«(cos v (1 + cosv)) / (1 – cos2 v) dv =âˆ« dx

â‡’ -âˆ«(cos v (1 + cos v)) / (sin2 v) dv = âˆ« dx

â‡’ -âˆ«(cot v cosec v + cot2 v) dv = âˆ« dx

â‡’ -âˆ« (cot v cosec v + cosec2 v – 1) dv = âˆ« dx

â‡’ -(-cosec v – cot v – v)= x + C

â‡’ cosec ( x – y ) + cot ( x – y ) + x – y = x + C

â‡’ cosec ( x – y ) + cot ( x – y ) – y = C

â‡’ ((1+cos ( x – y )) / sin ( x – y )) – y = C

â‡’ 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

â‡’ 1 – dy/dx = dv/dx

â‡’ dy/dx = 1 – dv/dx

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

â‡’ dv/dx = 1 – (v + 3)/ (2v + 5)

â‡’ dv/dx = (2v + 5 – v – 3)/ 2v + 5

â‡’ dv/dx = (v + 2) / (2v + 5)

â‡’ (2v + 5)/(v + 2)dv = dx

Integrating both sides, we get

âˆ«(2v + 5)/(v + 2) dv = âˆ«dx

â‡’ âˆ«(2v + 4 + 1)/(v + 2) dv = âˆ«dx

â‡’ âˆ«((2v + 4)/(v + 2) + 1/(v + 2))dv = âˆ«dx

â‡’ 2âˆ«dv + âˆ«1/(v + 2)dv = âˆ«dx

â‡’ 2v + log |v + 2| = x + C

â‡’ 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

â‡’ 1 + dy/dx = dv/dx

â‡’ dy/dx = dv/dx – 1

Therefore, dv/dx – 1 = v2

â‡’ dv/dx = v2 + 1

â‡’ 1/(v2 + 1) dv = dx

Integrating both sides, we get

âˆ«1/(v2 + 1) dv = âˆ«dx

â‡’ tan-1 v = x + C

â‡’ v = tan (x + C)

â‡’ x + y = tan (x + C)

### Question 5: (x + y)2 dy/dx = 1

Solution:

We have,

(x + y)2 dy/dx = 1

â‡’ dy/dx = 1/( x + y)2

Let x + y = v

â‡’ 1 + dy/dx = dv/dx

â‡’ dy/dx = dv/dx – 1

Therefore, dv/dx – 1 = 1/v2

â‡’ dv/dx = 1/v2 + 1

â‡’ v2/(v2 + 1) dv = dx

Integrating both sides, we get

âˆ«v2/(v2 + 1) dv = âˆ«dx

â‡’ âˆ«v2 + 1 – 1/(v2 + 1) dv = âˆ«dx

â‡’ âˆ«(1- 1/(v2 + 1) dv = âˆ«dx

â‡’ v – tan-1 v = x + C

â‡’ x + y – tan-1 (x + y) = x + C

â‡’ y – tan-1 (x + y) = C

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

Solution:

We have,

cos2 ( x – 2y ) = 1 – 2dy/dx

â‡’ 2dy/dx = 1 – cos2 (x – 2y)

Let x – 2y = v

â‡’ 1 – 2 dy/dx = dv/dx

â‡’ 2 dy/dx = 1 – dv/dx

Therefore, 1 – dv/dx = 1 – cos2 v

â‡’ dv/dx = cos2 v

â‡’ sec2 v dv = dx

Integrating both sides, we get

âˆ« sec2 v dv = âˆ«dx

â‡’ tan v = x – C

â‡’ tan (x – 2y) = x – C

â‡’ x = tan (x – 2y) + C

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

Solution:

We have,

dy/dx = sec(x + y)

â‡’ dy/dx = 1/cos ( x + y)

Let x + y = v

â‡’ 1 + dy/dx = dv/dx

â‡’ dy/dx = dv/dx -1

Therefore, dv/dx – 1 = 1/cos v

â‡’ dv/dx = (cos v + 1)/ cos v

â‡’ cos v/(cos v + 1) dv = dx

Integrating both sides, we get

âˆ« cos v/(cos v + 1) dv = âˆ« dx

â‡’ âˆ« cos v (1 – cos v)/(1 – cos2 v ) dv = âˆ« dx

â‡’ âˆ« cos v (1 – cos v)/sin2 v  dv = âˆ« dx

â‡’ âˆ« (cos v – cos^2 v)/sin2 v  dv = âˆ« dx

â‡’ âˆ«(cot v cosec v – cot2 v) dv = âˆ« dx

â‡’ âˆ«(cot v cosec v – cosec2 v + 1) dv = âˆ« dx

â‡’ – cosec v + cot v + v = x + C

â‡’ – cosec (x + y) + cot (x + y) + x + y = x + C

â‡’ – cosec (x + y) + cot (x + y) + y = C

â‡’ ((-1 + cos (x + y)) / sin (x + y)) + y = C

â‡’ – tan ((x + y)/2) + y = C

â‡’ 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

â‡’ dy/dx = dv/dx – 1

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

â‡’ dy/dx = sin v/cos v + 1

â‡’ dy/dx = (sin v + cos v)/ cos v

â‡’ cos v/(sin v + cos v) dv = dx

Integrating both sides, we get

â‡’ âˆ« cos v/(sin v + cos v) dv =âˆ« dx

â‡’ 1/2 âˆ« {(sin v + cos v) +  (cos v – sin v)}/(sin v + cos v) dv = âˆ«dx

â‡’ 1/2 âˆ« dv + 1/2 âˆ« (cos v – sin v) / (sin v + cos v) dv = âˆ«dx

â‡’ 1/2 v + 1/2 âˆ« (cos v – sin v)/(sin v + cos v) dv = x

Putting sin v + cos v = t

â‡’ (cos v – sin v) dv = dt

Therefore, 1/2 v + 1/2 âˆ« dt/t = x

â‡’ 1/2 v + 1/2 log |t| = x + C

â‡’ (x + y) + 1/2 log |sin (x + y) + cos (x + y)| = x + C

â‡’ 1/2 (y – x) + 1/2 log |sin (x + y) + cos (x + y)| = C

â‡’ (y – x) + log |sin (x + y) + cos (x + y)| = 2C

â‡’ 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

â‡’ x dx + y dx -x dy – y dy = dx + dy

â‡’ (x + y -1)dx = ( x + y +1) dy

â‡’ dy/dx = (x + y -1)/(x + y + 1)

Let x + y = v

Therefore, 1 + dy/dx =dv/dx

â‡’ dy/dx = dv/dx – 1

Therefore,  dv/dx -1 = (v – 1)/(v +1)

â‡’ dv/dx = (v – 1)/(v +1) + 1

â‡’ dv/dx = (v – 1 + v +1)/(v +1)

â‡’ dv/dx = 2v / (v +1)

â‡’ (v +1) / 2v dv = dx

Integrating both sides, we get

âˆ«(v + 1)/2v dv = âˆ« dx

â‡’ 1/2 âˆ«dv + 1/2 âˆ«1/v dv = âˆ«dx

â‡’ 1/2 v + 1/2 log |v| = x + C

â‡’ 1/2 (x + y) + 1/2 log |x + y| = x + C

â‡’ 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

â‡’ dy/dx = 1/(x + y + 1)

Let x + y + 1 = v

Therefore, 1 + dy/dx = dv/dx

â‡’ dy/dx = dv/dx – 1

Therefore, dv/dx – 1= 1/v

â‡’ dv/dx = 1/v + 1

â‡’ v/(v + 1) dv = dx

Integrating both sides, we get

âˆ« v/(v + 1) dv = âˆ« dx

â‡’ âˆ« (v + 1 – 1)/(v + 1) dv = âˆ« dx

â‡’ âˆ« (1 – 1/(v + 1))dv = âˆ« dx

â‡’ v – log |v + 1| = x + K

â‡’ x + y + 1 – log |x + y+ 1 + 1| = x + K

â‡’ y – log |x + y + 2| = K – 1

â‡’ y – log |x + y + 2| = C1        ( C1 = K – 1)

â‡’ y – C1 = log |x + y + 2|

â‡’ ey – C1 = x + y + 2

â‡’ ey / eC1 = x + y + 2

â‡’ e– C ey = x + y + 2

â‡’ C ey  = x + y + 2                  (C = e– C1)

â‡’ x =  C ey – y – 2

### Question 11: dy/dx + 1 = ex + y

Solution:

We have,

dy/dx + 1 = ex + y                      . . . (1)

Let x + y = t

â‡’ 1 + dy/dx = dt/dx

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

dt/dx = et

â‡’ e– t dt = dx

â‡’ – e– t = x + C

â‡’ – e– (x + y) = x + C         [Since t = x + y]

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