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NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3 - Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3 - Differential Equations

Edited By Komal Miglani | Updated on May 08, 2025 02:26 PM IST | #CBSE Class 12th

Consider a tank with a hole at the bottom. The exit rate of water is a function of the water level in the tank. As the water level reduces, the flow rate reduces. This can be represented with a differential equation.

The NCERT Solutions for Class 12 Maths Chapter 9 – Exercise 9.3 teach students exactly how to determine solutions to such equations. These are designed by subject matter experts from Careers360 as per the new CBSE 2025–26 syllabus and are described in depth in step-by-step manner. NCERT Chapter 9 Exercise 9.3 is differential equations under the category of variable separable, where the variables can be rearranged such that all the terms of one variable are placed on one side and the remaining terms are placed on the other side. Students will also encounter special cases where the equation needs to be transformed into separable form prior to using the method.

This Story also Contains
  1. Class 12 Maths Chapter 9 Exercise 9.3 Solutions: Download PDF
  2. NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.3
  3. Topics covered in Chapter 9 Differential Equation: Exercise 9.3
  4. NCERT Solutions Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

Class 12 Maths Chapter 9 Exercise 9.3 Solutions: Download PDF

This material provides easy solutions to all the questions of Exercise 9.3 of Differential Equations. The PDF can be downloaded by the students for practice and enhancement of the chapter for board and and competitive exams

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NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.3

Question 1: Find the general solution: dydx=1cosx1+cosx

Answer:

Given,

dydx=1cosx1+cosx

dydx=2sin2x22cos2x2=tan2x2

dy=(sec2x21)dx

dy=sec2x2dxdx

y=2tan1x2x+C

Question 2: Find the general solution: dydx=4y2 (2<y<2)

Answer:

Given, in the question

dydx=4y2

dy4y2=dxdy4y2=dx

(dya2y2=sin1ya)

The required general solution:

sin1y2=x+C

Question 3: Find the general solution: dydx+y=1(y1)

Answer:

Given, in the question

dydx+y=1

dydx=1y

dy1y=dx

(dxx=lnx)

log(1y)=x+C  (We can write C=logk)

logk(1y)=x

1y=1kex

The required general equation

y=11kex

Question 4: Find the general solution: sec2xtanydx+sec2ytanxdy=0

Answer:

Given,

sec2xtanydx+sec2ytanxdy=0

sec2ytanydy=sec2xtanxdx

sec2ytanydy=sec2xtanxdx

Now, let tany = t and tanx = u

sec2ydy=dt and sec2xdx=du

dtt=duu

logt=logu+logkt=1ku

tany=1ktanx

Question 5: Find the general solution:

(ex+ex)dy(exex)dx=0

Answer:

Given, in the question

(ex+ex)dy(exex)dx=0

dy=(exex)(ex+ex)dx

Let,

(ex+ex)=t

(exex)dx=dt

dy=dtt

y=logt+C

y=log(ex+ex)+C

This is the general solution

Question 6: Find the general solution: dydx=(1+x2)(1+y2)

Answer:

Given, in the question

dydx=(1+x2)(1+y2)

dy(1+y2)=(1+x2)dx

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

tan1y=x+x33+C

Question 7: Find the general solution: ylogydxxdy=0

Answer:

Given,

ylogydxxdy=0

1ylogydy=1xdx

let logy = t

=> 1/ydy = dt

dtt=1xdx

logt=logx+logk

t=kx

logy=kx

This is the general solution

Question 8: Find the general solution: x5dydx=y5

Answer:

Given, in the question

x5dydx=y5

dyy5=dxx5

y44=x44+C

1y4+1x4=C

This is the required general equation.

Question 9: Find the general solution: dydx=sin1x

Answer:

Given, in the question

dydx=sin1x

dy=sin1xdx

Now,

(u.v)dx=uvdx(dudx.vdx)dx

Here, u = sin1x and v = 1

y=sin1x.x(11x2.x)dx

Let 1x2=t

2xdx=dtxdx=dt/2

y=xsin1x+(dt2t)

y=xsin1x+12.2t+C

y=xsin1x+1x2+C

Question 10: Find the general solution extanydx+(1ex)sec2ydy=0

Answer:

Given,

extanydx+(1ex)sec2ydy=0

extanydx=(1ex)sec2ydy

sec2ytanydy=ex(1ex)dx

let tany=t and 1ex=u

sec2ydy=dt and exdx=du

dtt=duu

logt=logu+logk

t=ku

tany=k(1ex)

Question 11: Find a particular solution satisfying the given condition:

(x3+x2+x+1)dydx=2x2+x; y=1 when x=0

Answer:

Given, in the question

(x3+x2+x+1)dydx=2x2+xdy=2x2+x(x3+x2+x+1)dx(x3+x2+x+1)=(x+1)(x2+1)


Now,

2x2+x(x+1)(x2+1)=Ax+1+Bx+Cx2+12x2+x(x+1)(x2+1)=Ax2+A(Bx+C)(x+1)(x+1)(x2+1)2x2+x=Ax2+A+Bx+Cx+C2x2+x=(A+B)x2+(B+C)x+A+C


Now comparing the coefficients

A+B=2;B+C=1;A+C=0


Solving these:

A=12, B=32,C=12


Putting the values of A,B,C :

2x2+x(x+1)(x2+1)=121(x+1)+123x1x2+1


Therefore,

dy=121x+1dx+123x1x2+1dxy=12log(x+1)+32xx2+1dx12dxx2+1y=12log(x+1)+342xx2+1dx12tan1x

Question 12: Find a particular solution satisfying the given condition:

x(x21)dydx=1; y=0 when x=2

Answer:

Given, in the question x(x21)dydx=1

dy=dxx(x21)dy=dxx(x1)(x+1)


Let,

1x(x+1)(x1)=Ax+Bx+1+cx11x(x+1)(x1)=A(x1)(x+1)+B(x)(x1)+C(x)(x+1)x(x+1)(x1)1x(x+1)(x1)=(A+B+C)x2+(BC)xAx(x+1)(x1)


Now comparing the values of A,B,C

A+B+C=0;BC=0;A=1


Solving these:

B=12 and C=12


Now putting the values of A,B,C,

1x(x+1)(x1)=1x+12(1x+1)+12(1x1)dy=1xdx+12(1x+1)dx+12(1x1)dxy=logx+12log(x+1)+12log(x1)+logcy=12log[c2(x1)(x+1)x2}iii)


Given, y=0 when x=2

0=12log[c2(21)(2+1)4}log3c24=0


3c2=4


Therefore,

y=12log[4(x1)(x+1)3x2]y=12log[4(x21)3x2]

Question 13: Find a particular solution satisfying the given condition:

cos(dydx)=a (aR); y=1 when x=0

Answer:

Given,

cos(dydx)=a

dydx=cos1a

dy=cos1a dx

y=xcos1a+c

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

y=xcos1a+1

Question 14: Find a particular solution satisfying the given condition:

dydx=ytanx; y=1 when x=0

Answer:

Given,

dydx=ytanx

dyy=tanx dx

logy=logsecx+logk

y=ksecx

Now, y=1 when x =0

1 = ksec0

k = 1

Putting the vlue of k:

y = sec x

Question 15: Find the equation of a curve passing through the point (0, 0) and whose differential equation is y=exsinx .

Answer:

We first find the general solution of the given differential equation

Given,

y=exsinx

dy=exsinxdx

LetI=exsinxdx

I=sinx.ex(cosx.ex)dx

I=exsinx[excosx(sinx.ex)dx]

2I=ex(sinxcosx)

I=12ex(sinxcosx)

y=12ex(sinxcosx)+c

Now, Since the curve passes through (0,0)

y = 0 when x =0

0=12e0(sin0cos0)+c

c=12

Putting the value of c, we get:

y=12ex(sinxcosx)+12

2y1=ex(sinxcosx)

Question 16: For the differential equation xydydx=(x+2)(y+2) , find the solution curve passing through the point (1, –1).

Answer:

We first find the general solution of the given differential equation

Given,

xydydx=(x+2)(y+2)

yy+2dy=x+2xdx

(y+2)2y+2dy=(1+2x)dx

(12y+2)dy=(1+2x)dx

y2log(y+2)=x+2logx+C

Now, Since the curve passes through (1,-1)

y = -1 when x = 1

12log(1+2)=1+2log1+C

10=1+0+CC=2

Putting the value of C:

y2log(y+2)=x+2logx+2

yx+2=2logx(y+2)

Question 17: Find the equation of a curve passing through the point (0,2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

According to the question,

ydydx=x

ydy=xdx

y22=x22+c

Now, Since the curve passes through (0,-2).

x =0 and y = -2

(2)22=022+cc=2

Putting the value of c, we get

y22=x22+2y2=x2+4

Question 18: At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is y+3x+4

According to the question,

dydx=2(y+3x+4)

dyy+3=2dxx+4

log(y+3)=2log(x+4)+logk

(y+3)=k(x+4)2

Now, Since the curve passes through (-2,1)

x = -2 , y =1

(1+3)=k(2+4)2k=1

Putting the value of k, we get

y+3=(x+4)2

Question 19: The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Volume of a sphere, V=43πr3

Given, Rate of change is constant.

dVdt=c

ddt(43πr3)=c

d(43πr3)=cdt

43πr3=ct+k

Now, at t=0, r=3 and at t=3 , r =6

Putting these value:

43π(3)3=c(0)+kk=36π

Also,

43π(6)3=c(3)+36π

3c=252πc=84π

Putting the value of c and k:

43πr3=84πt+36π

r3=(21t+9)(3)=62t+27

r=62t+273

Question 20: In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to question,

dpdt=(r100)p

dpp=(r100)dt

logp=r100t+C

p=ert100+C

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

100=er(0)100+C=eC

Also,

, 200=er(10)100+C=er10.eC=er10.100

er10=2

r10=ln2=0.6931

r=6.93

So value of r = 6.93%

Question 21: In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e 0.5 = 1.648).

Answer:

Let p be the principal amount and t be the time.

According to question,

dpdt=(5100)p

dpp=(120)dt

logp=120t+C

p=et20+C

Now, at t =0 , p = 1000

Putting these values,

1000=e(0)20+C=eC

Also, At t=10 \,

p=e(10)20+C=e12.eC=e12.1000

p=(1.648)(1000)=1648

After 10 years, the total amount would be Rs.1648

Question 22: In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let n be the number of bacteria at any time t.

According to question,

dndt=kn  (k is a constant)

dnn=kdt

logn=kt+C

Now, at t=0, n = 100000

log(100000)=k(0)+C

C=5

Again, at t=2, n= 110000

log(110000)=k(2)+5

log11+4=2k+5

2k=log111=log1110

k=12log1110

Using these values, for n= 200000

log(200000)=kt+C

log2+5=kt+5(12log1110)t=log2

t=2log2log1110

Question 23: The general solution of the differential equation dydx=ex+y is

(A) ex+ey=C

(B) ex+ey=C

(C) ex+ey=C

(D) ex+ey=C

Answer:

Given,

dydx=ex+y

dydx=ex.eydyey=ex.dx

ey=ex+Cex+ey=K    (OptionA)

Also check -

Topics covered in Chapter 9 Differential Equation: Exercise 9.3


TopicsDescriptionExample
Variable Separable MethodInvolves separating variables x and y on opposite sides of the equation and integrating both sides. Used when a differential equation can be expressed as: dydx=g(x)h(y)

Solve: dydx=xy

1ydy=xdx1ydy=xdx

Equations Already in Separable FormSome differential equations are already arranged so that dy and dx can be directly separated. Solve: dydx=x21+y2(1+y2)dy=x2dx(1+y2)dy=x2dxy+y33=x33+C
Implicit SolutionsSolutions are sometimes left in implicit form (not solved for y explicitly). This is valid and acceptable in many cases.From above: y+y33=x33+C is an implicit solution.
Solving Using Initial ConditionsAfter finding the general solution, plug in the given values (initial conditions) to find the constant C for the particular solution.Given: dydx=xy,y=1 when x=0
General: y=Cex22
Apply: 1=Ce0C=1
Final: y=ex22
Rearranging to Separable FormIf the equation isn't already in separable form, rearrange terms algebraically to isolate variables on each side.

Solve: xdydx=y

dydx=yx1ydy=1xdx

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Frequently Asked Questions (FAQs)

1. What is the order and degree of differential equations handled in topic 9.1?

The differential equations of order 1 and degree 1 are handled.

2. What type of differential equation is solved in exercise 9.3 Class 12 Maths?

Differential equations of variable separable type are solved in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.3.

3. How many solved examples are given in topic 9.5.1?

Six example problems and their solutions are given under topic 9.5.1.

4. How many questions are there in Class 12 Maths chapter 9 exercise 9.3?

Twenty-three questions are explained in the Class 12th Maths chapter 6 exercise 9.3.

5. How many questions of exercise 9.1 are multiple-choice questions?

One question is of objective type with 4 choices.

6. In topic 9.4, the number of sub-topics discussed is?

Three subtopics are discussed in topic 9.4.

7. In session 9.4 of NCERT Class, 12 Maths book how many methods to solve the first-degree first-order differential equations are discussed?

3 methods are discussed. 

8. Name the subtopics discussed in 9.4

The topics discussed are listed below

Differential equations of variable separable type, homogeneous and linear differential equations. 

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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