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NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5 - Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5 - Differential Equations

Edited By Komal Miglani | Updated on May 08, 2025 02:24 PM IST | #CBSE Class 12th

Suppose we have the example of draining out water from a tank, where the drainage rate is directly proportional to the remaining water in the tank. This situation can be represented by a first-order linear differential equation, which can be used to forecast the time taken for the tank to empty, as well as the quantity of water left after a specified time. Prior to this exercise, the NCERT book has already presented solved examples that lead students through the step-by-step solving of such equations—so it is simple to solve the problems of NCERT Exercise 9.5.

The NCERT solutions, designed by experienced teachers, follow the CBSE syllabus for the 2025-26 session and are a precious asset for board exams as well as competitive exams such as the JEE Main. The solutions are presented in a step-by-step and logical format that is easy to comprehend, confidence-building, and enhances analytical ability. Through acquiring these skills, learners are able to achieve proficiency in determining linear differential equations, in solving them using the integrating factor method and comprehending general and specific solutions correctly.

This Story also Contains
  1. Class 12 Maths Chapter 9 Exercise 9.5 Solutions: Download PDF
  2. NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.5
  3. Topics covered in Chapter 9 Differential Equation: Exercise 9.5
  4. NCERT Solutions Subject Wise
  5. Subject Wise NCERT Exemplar Solutions
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The CBSE Class 12th supplementary exam will be conducted in the first or second week of July 2025, and it will be held on the same syllabus as the main examinations 2025 were conducted. 

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Class 12 Maths Chapter 9 Exercise 9.5 Solutions: Download PDF

This material provides easy solutions to all the questions of Exercise 9.5 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 9.5

Question 1: Find the general solution:

dydx+2y=sinx

Answer:

Given equation is
dydx+2y=sinx
This is dydx+py=Q type where p = 2 and Q = sin x
Now,
I.F.=epdx=e2dx=e2x
Now, the solution of given differential equation is given by relation
Y(I.F.)=(Q×I.F.)dx+C
Y(e2x)=(sinx×e2x)dx+C
Let I=(sinx×e2x)
I=sinxe2xdx(d(sinx)dx.e2xdx)dx

I=sinx.e2x2(cosx.e2x2)

I=sinx.e2x212(cosxe2xdx(d(cosx)dx.e2xdx))dx

I=sinxe2x212(cosx.e2x2+(sinx.e2x2))

I=sinxe2x212(cosx.e2x2+I2)           (I=sinxe2x)5I4=e2x4(2sinxcosx)

I=e2x5(2sinxcosx)
Put the value of I in our equation
Now, our equation become
Y.ex2=e2x5(2sinxcosx)+C
Y=15(2sinxcosx)+C.e2x
Therefore, the general solution is Y=15(2sinxcosx)+C.e2x

Question 2: Solve for general solution:

dydx+3y=e2x

Answer:

Given equation is
dydx+3y=e2x
This is dydx+py=Q type where p = 3 and Q=e2x
Now,
I.F.=epdx=e3dx=e3x
Now, the solution of given differential equation is given by the relation
Y(I.F.)=(Q×I.F.)dx+C
Y(e3x)=(e2x×e3x)dx+C
Y(e3x)=(ex)dx+C

Y(e3x)=ex+C

Y=e2x+Ce3x
Therefore, the general solution is Y=e2x+Ce3x

Question 3: Find the general solution

dydx+yx=x2

Answer:

Given equation is
dydx+yx=x2
This is dydx+py=Q type where p=1x and Q=x2
Now,
I.F.=epdx=e1xdx=elogx=x
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(x)=(x2×x)dx+C
y(x)=(x3)dx+C

y.x=x44+C
Therefore, the general solution is yx=x44+C

Question 4: Solve for General Solution.

dydx+(secx)y=tanx (0x<π2)

Answer:

Given equation is
dydx+(secx)y=tanx (0x<π2)
This is dydx+py=Q type where p=secx and Q=tanx
Now,
I.F.=epdx=esecxdx=elog|secx+tanx|=secx+tanx (0xπ2secx>0,tanx>0)
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(secx+tanx)=((secx+tanx)×tanx)dx+C
y(secx+tanx)=(secxtanx+tan2x)dx+C

y(secx+tanx)=secx+(sec2x1)dx+C

y(secx+tanx)=secx+tanxx+C
Therefore, the general solution is y(secx+tanx)=secx+tanxx+C

Question 5: Find the general solution.

cos2xdydx+y=tanx(0x<π2)

Answer:

Given equation is
cos2xdydx+y=tanx(0x<π2)
we can rewrite it as
dydx+sec2xy=sec2xtanx
This is dydx+py=Q where p=sec2x and Q=sec2xtanx
Now,
I.F.=epdx=esec2xdx=etanx
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(etanx)=((sec2xtanx)×etanx)dx+C
yetanx=sec2xtanxetanxdx+C
take
etanx=t

sec2x.etanxdx=dt
t.logtdt=logt.tdt(d(logt)dt.tdt)dt

t.logtdt=logt.t22(1t.t22)dt

t.logtdt=logt.t22t2dt

t.logtdt=logt.t22t24

t.logtdt=t24(2logt1)
Now put again t=etanx
sec2xtanxetanxdx=e2tanx4(2tanx1)
Put this value in our equation

yetanx=e2tanx4(2tanx1)+C
Therefore, the general solution is y=etanx4(2tanx1)+Cetanx

Question 6: Solve for General Solution.

xdydx+2y=x2logx

Answer:

Given equation is
xdydx+2y=x2logx
Wr can rewrite it as
dydx+2.yx=xlogx
This is dydx+py=Q type where p=2x and Q=xlogx
Now,
I.F.=epdx=e2xdx=e2logx=elogx2=x2 (0xπ2secx>0,tanx>0)
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(x2)=(xlogx×x2)dx+C
x2y=x3logx+C
Let
I=x3logxI=logxx3dx(d(logx)dx.x3dx)dx

I=logx.x44(1x.x44)dx

I=logx.x44(x34)dx

I=logx.x44x416
Put this value in our equation
x2y=logx.x44x416+C

y=x216(4logx1)+C.x2
Therefore, the general solution is y=x216(4logx1)+C.x2

Question 7: Solve for general solutions.

xlogxdydx+y=2xlogx

Answer:

Given equation is
xlogxdydx+y=2xlogx
we can rewrite it as
dydx+yxlogx=2x2
This is dydx+py=Q type where p=1xlogx and Q=2x2
Now,
I.F.=epdx=e1xlogxdx=elog(logx)=logx
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(logx)=((2x2)×logx)dx+C

take
I=((2x2)×logx)dx
I=logx.2x2dx(d(logx)dt.x22dx)dx

I=logx.2x+(1x.2x)dx

I=logx.2x+2x2dx

I=logx.2x2x
Put this value in our equation

ylogx=2x(logx+1)+C
Therefore, the general solution is ylogx=2x(logx+1)+C

Question 8: Find the general solution.

(1+x2)dy+2xydx=cotxdx (x0)

Answer:

Given equation is
(1+x2)dy+2xydx=cotxdx (x0)
we can rewrite it as
dydx+2xy(1+x2)=cotx1+x2
This is dydx+py=Q type where p=2x1+x2 and Q=cotx1+x2
Now,
I.F.=epdx=e2x1+x2dx=elog(1+x2)=1+x2
Now, the solution of the given differential equation is given by the relation
y(I.F.)=(Q×I.F.)dx+C
y(1+x2)=((cotx1+x2)×(1+x2))dx+C
y(1+x2)=cotxdx+C

y(1+x2)=log|sinx|+C

y=(1+x2)1log|sinx|+C(1+x2)1
Therefore, the general solution is y=(1+x2)1log|sinx|+C(1+x2)1

Question 9: Solve for general solution.

xdydx+yx+xycotx=0 (x0)

Answer:

Given equation is
xdydx+yx+xycotx=0 (x0)
we can rewrite it as
dydx+y.(1x+cotx)=1
This is dydx+py=Q type where p=(1x+cotx) and Q=1
Now,
I.F.=epdx=e(1x+cotx)dx=elogx+log|sinx|=x.sinx
Now, the solution of the given differential equation is given by the relation
y(I.F.)=(Q×I.F.)dx+C
y(x.sinx)=1×xsinxdx+C
y(x.sinx)=xsinxdx+C
Lets take
I=xsinxdx

I=x.sinxdx(d(x)dx.sinxdx)dx

I=x.cosx+(cosx)dx

I=xcosx+sinx
Put this value in our equation
y(x.sinx)=xcosx+sinx+C

y=cotx+1x+Cxsinx
Therefore, the general solution is y=cotx+1x+Cxsinx

Question 10: Find the general solution.

(x+y)dydx=1

Answer:

Given equation is
(x+y)dydx=1
we can rewrite it as
dydx=1x+y

x+y=dxdydxdyx=y
This is dxdy+px=Q type where p=1 and Q=y
Now,
I.F.=epdy=e1dy=ey
Now, the solution of given differential equation is given by relation
x(I.F.)=(Q×I.F.)dy+C
x(ey)=y×eydy+C
xey=y.eydy+C
Lets take
I=yeydy

I=y.eydy(d(y)dy.eydy)dy

I=y.ey+eydyI=yeyey
Put this value in our equation
x.ey=ey(y+1)+C

x=(y+1)+Cey

x+y+1=Cey
Therefore, the general solution is x+y+1=Cey

Question 11: Solve for general solution.

ydx+(xy2)dy=0

Answer:

Given equation is
ydx+(xy2)dy=0
we can rewrite it as
dxdy+xy=y
This is dxdy+px=Q type where p=1y and Q=y
Now,
I.F.=epdy=e1ydy=elogy=y
Now, the solution of given differential equation is given by relation
x(I.F.)=(Q×I.F.)dy+C
x(y)=y×ydy+C
xy=y2dy+C
xy=y33+C
x=y23+Cy
Therefore, the general solution is x=y23+Cy

Question 12: Find the general solution.

(x+3y2)dydx=y (y>0)

Answer:

Given equation is
(x+3y2)dydx=y (y>0)
we can rewrite it as
dxdyxy=3y
This is dxdy+px=Q type where p=1y and Q=3y
Now,
I.F.=epdy=e1ydy=elogy=y1=1y
Now, the solution of given differential equation is given by relation
x(I.F.)=(Q×I.F.)dy+C
x(1y)=3y×1ydy+C
xy=3dy+C
xy=3y+C
x=3y2+Cy
Therefore, the general solution is x=3y2+Cy

Question 13: Solve for particular solution.

dydx+2ytanx=sinx; y=0 when x=π3

Answer:

Given equation is
dydx+2ytanx=sinx; y=0 when x=π3
This is dydx+py=Q type where p=2tanx and Q=sinx
Now,
I.F.=epdx=e2tanxdx=e2log|secx|=sec2x
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(sec2x)=((sinx)×sec2x)dx+C
y(sec2x)=(sin×1cosx×secx)dx+C

y(sec2x)=tanxsecxdx+C

y.sec2x=secx+C
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x=π3
at x=π3
secπ3=secπ3+C

C=2
Now,

y.sec2x=secx2

ycos2x=1cosx2

y=cosx2cos2x
Therefore, the particular solution is y=cosx2cos2x

Question 14: Solve for particular solution.

(1+x2)dydx+2xy=11+x2; y=0 when x=1

Answer:

Given equation is
(1+x2)dydx+2xy=11+x2; y=0 when x=1
we can rewrite it as
dydx+2xy1+x2=1(1+x2)2
This is dydx+py=Q type where p=2x1+x2 and Q=1(1+x2)2
Now,
I.F.=epdx=e2x1+x2dx=elog|1+x2|=1+x2
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(1+x2)=(1(1+x2)2×(1+x2))dx+C
y(1+x2)=1(1+x2)dx+C

y(1+x2)=tan1x+C
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
0.(1+12)=tan11+C

C=π4
Now,
y(1+x2)=tan1xπ4
Therefore, the particular solution is y(1+x2)=tan1xπ4

Question 15: Find the particular solution.

dydx3ycotx=sin2x; y=2 when x=π2

Answer:

Given equation is
dydx3ycotx=sin2x; y=2 when x=π2
This is dydx+py=Q type where p=3cotx and Q=sin2x
Now,
I.F.=epdx=e3cotxdx=e3log|sinx|=sin3x=1sin3x
Now, the solution of given differential equation is given by relation
y(I.F.)=(Q×I.F.)dx+C
y(1sin3x)=(sin2x×1sin3x)dx+C
ysin3x=(2sinxcosx×1sin3x)dx+C
ysin3x=(2×cosxsinx×1sinx)dx+C
ysin3x=(2×cotx×cosecx)dx+C
ysin3x=2cosecx+C
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when x=π2
at x=π2
2sin3π2=2cosecπ2+C

2=2+CC=4
Now,
y=4sin3x2sin2x
Therefore, the particular solution is y=4sin3x2sin2x

Question 16: Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

Let f(x , y) is the curve passing through origin
Then, the slope of tangent to the curve at point (x , y) is given by dydx
Now, it is given that
dydx=y+x

dydxy=x
It is dydx+py=Q type of equation where p=1 and Q=x
Now,
I.F.=epdx=e1dx=ex
Now,
y(I.F.)=(Q×I.F.)dx+C
y(ex)=(x×ex)dx+C
Now, Let
I=(x×ex)dx

I=x.exdx(d(x)dx.exdx)dx

I=xex+exdx

I=xexex

I=ex(x+1)
Put this value in our equation
yex=ex(x+1)+C
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
0.e0=e0(0+1)+C

C=1
Our final equation becomes
yex=ex(x+1)+1

y+x+1=ex
Therefore, the required equation of the curve is y+x+1=ex

Question 17: Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let f(x , y) is the curve passing through point (0 , 2)
Then, the slope of tangent to the curve at point (x , y) is given by dydx
Now, it is given that
dydx+5=y+x

dydxy=x5
It is dydx+py=Q type of equation where p=1 and Q=x5
Now,
I.F.=epdx=e1dx=ex
Now,
y(I.F.)=(Q×I.F.)dx+C
y(ex)=((x5)×ex)dx+C
Now, Let
I=((x5)×ex)dx

I=(x5).exdx(d(x5)dx.exdx)dx

I=(x5)ex+exdx

I=xexex+5ex

I=ex(x4)
Put this value in our equation
yex=ex(x4)+C
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
2.e0=e0(04)+C

C=2
Our final equation becomes
yex=ex(x4)2

y=4x2ex
Therefore, the required equation of curve is y=4x2ex

Question 18: The Integrating Factor of the differential equation xdydxy=2x2 is

(A) ex

(B) ey

(C) 1x

(D) x

Answer:

Given equation is
xdydxy=2x2
we can rewrite it as
dydxyx=2x
Now,
It is dydx+py=Q type of equation where p=1x and Q=2x
Now,
I.F.=epdx=e1xdx=elogx=x1=1x
Therefore, the correct answer is (C)

Question 19: The Integrating Factor of the differential equation (1y2)dxdy+yx=ay  (1<y<1) is

(A) 1y21

(B) 1y21

(C) 11y2

(D) 11y2

Answer:

Given equation is
(1y2)dxdy+yx=ay  (1<y<1)
we can rewrite it as
dxdy+yx1y2=ay1y2
It is dxdy+px=Q type of equation where p=y1y2 and Q=ay1y2
Now,
I.F.=epdy=ey1y2dy=elog|1y2|2=(1y2)12=11y2
Therefore, the correct answer is (D)

Topics covered in Chapter 9 Differential Equation: Exercise 9.5

TopicsDescriptionExample
Identifying Linear Differential EquationEquation in the form: dydx+Py=Qdydx+3y=6
Finding the Integrating Factor (IF)Use: IF=ePdx, where P is the coefficient of yIF=e3dx=e3x
Multiplying both sides with the IFMultiply full equation by IF to simplify into exact derivativee3xdydx+3e3xy=6e3x
Expressing as a derivativeLeft side becomes the derivative of y×IFddx(ye3x)=6e3x
Integrating both sidesSolve the equation by integrating both sidesye3x=6e3xdx=2e3x+C
Finding the General SolutionSolve for y from the integrated equationy=2+Ce3x
Finding Particular Solution (if values given)Use given values to calculate the constant CIf y=5 when x=0:5=2+CC=3

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NCERT Solutions Subject Wise

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. How many questions are explained in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.5?

19 questions are given in Exercise 9.5, Class 12 Maths.

2. Which topic is projected in Class 12 Maths chapter 9 exercise 9.5?

Linear differential equations have been discussed through exercise 9.5.

3. How many multiple-choice questions are given in the Class 12th Maths chapter 9 exercise 9.5?

Two questions of 4 choices are given in exercise 9.5.

4. What is the topic discussed after Class 12th Maths chapter 9 exercise 9.5?

No topics are discussed after exercise 9.5.

5. Is there any exercise after 9.5?

Yes, miscellaneous exercise is given after 9.5.

6. Is it mandatory to solve exercise 9.5 Class 12 Maths for board exam?

The concepts covered in NCERT solutions for Class 12 Maths chapter 9 exercise 9.5 are important from the chapter differential equations.

7. Can I skip the chapter differential equations for board exams?

No, it is important to go through the chapter as it holds a good weightage for board exams

8. What is the number of practice exercises given in the chapter differential equations?

6 exercises are present in the Class 12 NCERT Maths chapter 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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