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RD Sharma Class 12 Exercise 21.8 Differential Equation Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 21.8 Differential Equation Solutions Maths - Download PDF Free Online

Updated on Jan 24, 2022 04:20 PM IST

The class 12 students need much focus and guidance in each subject while preparing for their exams. Jumping from one solution book to another for each topic or concept would be hectic. Especially when mathematics comes into play, the students feel hard to solve the sums in the Differential Equations concept. Authorized reference materials like the RD Sharma Class 12th Exercise 21.8 books are a boon for these students.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
  2. Differential Equations Excercise: 21.8
  3. RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.8

Differential equations exercise 21.8 question 1

Answer: tan1(x+y+1)=x+c
Given:dydx=(x+y+1)2

To find :- solve the differential equation.

Hint :- differential equation of the form

dydx=(ax+by+c)can be reduced to variable separable form by the substitution ax+by+c=v

Solution:-we have

dydx=(x+y+1)2 ........(i)

Letx+y+1=v

Differentiating with respect to x, we get

1+dydx=dvdxdydx=dvdx1 ....(ii)

Now, substituting equation (ii) in equation (i), we get,
dvdx1=(x+y+1)2dvdx1=v2(x+y+1=v)dvdx=v2+1dvv2+1=dx( taking like variable on same side)
Integrating on both sides, we get,
dvv2+1=dx
tan1v=x+c (dxa2+x2=1atan1(xa))
Putting the value of V, we get,
tan1(x+y+1)=x+c , which is the required solution.

Differential equations exercise 21.8 question 2

Answer: cotxy2=y+C1
Gievn:-dydxcot(xy)=1

To find :- solve the differential equation.

Hint :- We will reduce the differential equation to variable separable form by substitution .

Solution ;- We have,

dydxcot(xy)=1 .....(i)

Let xy=v

Differentiating with respect to x, we get,

ddx(xy)=ddx(v)

1dydx=dvdx

dydx=1dvdx (transposing)

Putting the value of dydx in Equ. (i), we get

(1dvdx)cosv=1(xy=v)1dvdx=1cosv1dvdx=secvdvdx=1secv

Now, taking like variable in same side, we get,

dv1secv=dxdx=dv1secvdx=(cosvcosv1)dv

Integrating in both side, we get,

dx=(cosv1cosv)dvx=cosv1Cosv×1+cosv1+cosvdvx=cosv+cos2v1cos2vdvx=(cosv+cos2vsin2v)dvx=(cosvsin2v+cos2vsin2v)dvx=(cotvCosecv+cot2)dv

x=(cotvcosecv+cosec2v1)dvx=(Cosecvcotvv)+Cx=Cosecv+cotv+v+C1x=1sinv+cosvsinv+v+C1xvC1=1+cosvsinv

xvC1=2los2v228 s2Tsinq2xvC1=cotv2 (put C=C1 and using 1+Cos2x=2Cos2x;sinx=2sinx2cosx2)

Putting v=xy

x/x+/y+C1=cot(xy2)

cot(xy2)=y+C1

(This is the required solution)

Differential equations exercise 21.8 question 3

Answer: 2(xy)+log(xy+2)=x+C
Given:dydx=(xy)+32(xy)+5
Hint : - first, we will separate the variables and then solve .
Solution:dydx=(xy)+32(xy)+5
Let xy=v
Differentiating with respect to x , we get,
ddx(xy)=dvdx1dydx=dvdxdydx=1dvdx
Putting dydx=1dvdx and xy=v in equation (i), we get,
1dvdx=v+32v+5dvdx=1(v+32v+5)dvdx=2v+5v32v+5dvdx=v+22v+5
Taking like variables in the same side,
2v+5v+2dv=dx
Now, integrating in both sides, we get,
2v+5v+2dv=dx
(2v+4+1v+2)dv=dx(2(v+2)v+2+1v+2)dv=dx(2+1v+2)dv=dx2v+log|v+2|=x+c
Putting v=xy
2(xy)+log|xy+2|=x+c
( this is the required solution).

Differential equations exercise 21.8 question 4

Answer:- x+y=tan(x+c)
Given:dydx=(x+y)2
Hint : - Differential equation of the formdydx=(ax+by+c)can be reduced to variable separable form by substitution ax+by+c=v

Solution : - We have,

dydx=(x+y)2

Let x+y=v

Differentiating with respect to x, we get,

ddx(x+y)=ddx(v)1+dydx=dvdxdydx=dvdx1

Putting the value of dydx in equation (i), we get

dvdx1=(x+y)2

dvdx1=v2 (x+y=v)

dvdx=v2+1

dvv2+1=dx
Integrating on both the sides, we get,

dvv2+1=dx

tan1v=x+c (dx2x2+1=tan1x)

Putting v=x+y, we get

tan1(x+y)=x+c

x+y=tan(x+c)

(This is the required solution).

Differential equations exercise 21.8 question 5

Answer: ytan1(x+y)=c
Given:(x+y)2dydx=1

Hint : - Use variable separable form by substitution.

Solution : - We have,

(x+y)2dydx=1 ......(i)

Let x+y=v

Differentiating with respect to x, we get,

ddx(x+y)=dvdx

1+dydx=dvdx

dydx=dvdx1 .....(ii)

Now, substituting equation (ii) in equation (i), we get

(x+y)2(dvdx1)=1

v2(dvdx1)=1

dvdx1=1v2

dvdx=1v2+1

dvdx=1+v2v2

Taking like variables on same side, we get,

v21+v2dv=dx

Integrating on both the sides, we get,

(v21+v2)dv=dx

(v2+111+v2)dv=dx

(v2+1v2+11v2+1)dv=dx

(11v2+1)dv=dx

dv1v2+1dv=dx

vtan1v=x+c((1x2+1)dx=tan1x)

Putting v=x+y, we have

(x+y)tan1(x+y)=x+c

ytan1(x+y)=c

(This is the required solution).

Differential equations exercise 21.8 question 6

Answer:x=tan(x2y)+c
Given:cos2(x2y)=12dydx
Hint : - first, we will separate variables and then solve.
Solution : - We have,

cos2(x2y)=12dydx .....(i)

Let x2y=v
Differentiating with respect to x, we get,

ddx(x2y)=dvdx

12dydx=dvdx

2dydx=1dvdx
Now, substituting equation (ii) in equation (i), we get

cos2(x2y)=1(1dvdx)cos2v=11+dvdx(v=x2y)cos2v=dvdx
Taking like variables on same side, we get,

dx=dvcos2v

dx=sec2vdv
Integrating on both the sides, we get,

dx=sec2vdv

x=tan(x2y)+c
(This is the required solution).

Differential equations exercise 21.8 question 7

Answer: y=tan(x+y2)+c
Given:dydx=sec(x+y)

Hint : - first, we will separate variables and then solve.

Solution : - We have,

dydx=sec(x+y)

Let x+y=v

Differentiating with respect to x, we get,

ddx(x+y)=dvdx1+dydx=dvdxdydx=dvdx1

Putting x + y = v and dydx = dvdx -1 in the given differential equation, we get,

dvdx1=secv

dvdx1=1cosv

dvdx=cosv+1cosv

Taking like variables on same side, we get,

cosvcosv+1 dv=dx

cosv(1cosv)(1+cosv)(1cosv)dv=dx

cosv(1cosv)1cos2vdv=dx

cosv(1cosv)sin2vdv=dx

Integrating on both the sides, we get,

cosv(1cosv)sin2vdv=dx

(cosvsin2vcos2vsin2v)dv=dx

(cotvcosecvcot2v)dv=dx

(cotvcosecvdv(cosec2v1)dv=dx(usingcot2v=cosec2v1)

cosecv+cotv+v)=x+c

Putting v=x+y,we have

cosec(x+y)+cot(x+y)+x+y=x+c

cosec(x+y)+cot(x+y)+y=c

1sin(x+y)+cos(x+y)sin(x+y)+y=c

(1cos(x+y)sin(x+y))+y=c

2sin2x+y22cosx+y2sinx+y2+y=c

tan(x+y)2+y=c

y=tan(x+y)2+c

(This is the required solution)

Differential equations exercise 21.8 question 8

Answer: yx+log|sin(x+y)+cos(x+y)|=c

Given:dydx=tan(x+y)

Hint : - first, we will separate variables and then solve.

Solution : - We have,

dydx=tan(x+y)

Letx+y=v

Differentiating with respect to x, we get,

ddx(x+y)=dvdx

1+dydx=dvdx

dydx=dvdx1

Putting x+y=v and dydx=dvdx1 in the given differential equation, we get,

dydx=dvdx1=tanv

dvdx=1+tanv

Taking like variables on same side, we get,

dv1+tanv=dx(11+sinvcosv)dv=dx(cosvcosv+sinv)dv=dx

Multiplying 2 on both the sides, we get,

(2cosvcosv+sinv)dv=2dx(1+cosvsinvcosv+sinv)dv=2dx

Integrating on both sides, we get,

(1+cosvsinvcosv+sinv)dv=2 dxdv+cosvsinvcosv+sinv dv=2 dxv+log|cosv+sinv|=2x+c

Putting v = x + y, we get,

(x+y)+logcos(x+y)+sin(x+y)=2x+cyx+log|cos(x+y)+sin(x+y)|=c

(This is the required solution).

Differential equations exercise 21.8 question 9

Answer: 12(yx)+12log|x+y|=c
Given:(x+y)(dxdy)=dx+dy

Hint : - first, we will separate variables and then solve.

Solution : - We have,

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

xdxxdy+ydxydy=dx+dy

xdx+ydxdx=dy+xdy+ydy

(x+y1)dx=(1+x+y)dy

dydx=x+y1x+y+1

Let x+y=v

Differentiating with respect to x, we get,

ddx(x+y)=dvdx

1+dydx=dvdx

dydx=dvdx1

Substituting (ii) in equation (i), we get,

dvdx1=x+y1x+y+1

dvdx1=v1v+1[x+y=v]

dvdx=v1v+1+1

dvdx=v1+v+1v+1

dvdx=2vv+1

Taking like variables on the same side, we get,

v+12vdv=dx

Integrating on both sides, we get,

v+12vdv=dx12(1+1v)dv=dx12(v+log|v|)=x+c
Putting v = x + y, we get,
12(x+y+log|x+y|)=x+c(x+y+log|x+y|)=2(x+c)(x+y+log|x+y|)=2x+2cyx+log|x+y|=2c12(yx)+12log|x+y|=c
(This is the required solution).

Differential equations exercise 21.8 question 10

Answer: x=ceyy2
Given:(x+y+1)dydx=1
Hint : - Differential equation of the form dydx=(ax+by+c)can be reduced to variable separable form by substitution ax+by+c=v
Solution : - We have,
(x+y+1)dydx=1 ........(i)
Let x+y=v
Differentiating with respect to x, we get,
ddx(x+y)dvdx
1+ddx=dvdx
ddx=dvdx1

Putting dydx = dvdx -1 and x + y = v in equation (i), we get,

(v+1)(dvdx1)=1(v+1)dvdx(v+1)=1(v+1)dvdx=1+1+v(v+1)dvdx=2+v

Taking like variables on the same side, we get,

v+1v+2dv=dx

Integrating on both sides, we get,

v+1v+2dv=dx

(11v+2)dv=dx

(vlog|v+2|)=x+logc1

Putting v = x + y, we get,

x+ylog|x+y+2|=x+logc1

x+yx=log|x+y+2|+logc1

y=logc1|x+y+2|

ey=c1(x+y+2)(c=1c1)

x=ceyy2
(This is the required solution).

Differential equations exercise 21.8 question 11

Answer: e(x+y)=x+c
Given: dydx+1=ex+y
Hint: Differential equation of the form dydx=(ax+by+c) can be reduced to variable separable form by substitution ax+by+c=v
Solution: We have,
dydx+1=ex+y .......(i)
Let x+y=v
Differentiating with respect to x, we get,
ddx(x+y)=dvdx
1+dydx=dvdx (ii) 
Substituting (ii) in equation (i), we get,
dvdx=ex+y
dvdx=ev[x+y=v]
Taking like variables on the same side, we get,
dvev=dx
evdv=dx
Integrating on both sides, we get,
evdv=dx
ev=x+c
Putting v=x+y, we get,
e(x+y)=x+c

(This is the required solution).


The class 12 mathematics portion consists of Differential Equations as its 21st chapter. There are around eleven exercises in the chapter, mostly revolving around the same concept in different methods. The eighth exercise, ex 21.8, has eleven questions present in the textbook. Again, the concept of this exercise is to solve the given differential equations. The sums are given under Level 1, and there is no Level 2 part. The students can use the RD Sharma Class 12 Chapter 21 Exercise 21.8 reference material for answer verification and clarify the other doubts they possess.

Even though the chapter relies on the same concept for the sums, there are differences in how each equation is being solved. Therefore, students need to concentrate on identifying the proper method to be solved for every equation. If a doubt strikes them, they can very well use the RD Sharma Class 12th Exercise 21.8 for reference. The CBSE students find RD Sharma books more optimal as the solutions follow the NCERT pattern.

Experts and teachers from the mathematics field have contributed their time and knowledge to this set of best solution books. The Class 12 RD Sharma Chapter 21 Exercise 21.8 Solution contains numerous sets of practice questions. When students practice more differential equation sums, they tend to score more marks than the others. To learn this concept effortlessly is a must for the students to have the RD Sharma Class 12 Solutions Differential Equation Ex 21.8 by their side.

This book provides an in-depth knowledge of the concept and makes the students skilled enough to compete in the public exams. And the benefit that excites most of the students is that the RD Sharma Class 12th Exercise 21.8 book can be downloaded without any payment from the Career 360 website. As the best solutions are available for free, why would anyone wish to prefer other sources? Therefore, download the RD Sharma solution materials that you require from the Career 360 website.

Another significant advantage is that the practice questions in the RD Sharma Class 12 Solutions Chapter 21 Ex 21.8 can be asked in the public exams. Hence, preparing with this set of books from day one will help the students score marks effortlessly.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. Why are RD Sharma solution books so popular?

Many schools recommend the RD Sharma solutions books due to the clarity and accuracy of answer keys.

2. Which books can the class 12 CBSE students use to clarify their doubts in mathematics, chapter 21?

The CBSE students can use the RD Sharma Class 12th Exercise 21.8 to clarify their doubts in this chapter.

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