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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:26 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5

NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9 Differential Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 9.5 Class 12 Maths chapter 9 comes under the topic of homogeneous differential equations. Before going to the sample questions and exercise 9.5 Class 12 Maths, the NCERT Book explains what is a homogeneous equation is and how to identify it. Then a few examples are given and proceed to NCERT solutions for Class 12 Maths chapter 9 exercise 9.5. Questions based on homogeneous differential equations are given in the Class 12 Maths chapter 9 exercise 9.5. Questions to find both particular and general solutions of homogeneous differential equations are present in the Class 12th Maths chapter 6 exercise 9.5.

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  1. NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5
  2. Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5
  3. More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5
  4. Also Read| Differential Equations Class 12th Notes
  5. Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5
  6. Key Features Of NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9
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  8. NCERT Solutions Subject Wise
  9. Subject Wise NCERT Exemplar Solutions

12th class Maths exercise 9.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

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Differential Equations Class 12 Chapter 9 Exercise: 9.5

Question:1 Show that the given differential equation is homogeneous and solve each of them. (x^2 + xy)dy = (x^2 + y^2)dx

Answer:

The given diffrential eq can be written as
\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}
Let F(x,y)=\frac{x^{2}+y^{2}}{x^{2}+xy}
Now, F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}
=\frac{x^{2}+y^{2}}{x^{2}+xy} = \lambda ^{0}F(x,y) Hence, it is a homogeneous equation.

To solve it put y = vx
Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v +x\frac{dv}{dx} = \frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ v +x\frac{dv}{dx} = \frac{1+v^{2}}{1+v}

x\frac{dv}{dx} = \frac{(1+v^{2})-v(1+v)}{1+v} = \frac{1-v}{1+v}

( \frac{1+v}{1-v})dv = \frac{dx}{x}

( \frac{2}{1-v}-1)dv = \frac{dx}{x}
Integrating on both side, we get;
\\-2\log(1-v)-v=\log x -\log k\\ v= -2\log (1-v)-\log x+\log k\\ v= \log\frac{k}{x(1-v)^{2}}\\
Again substitute the value y = \frac{v}{x} ,we get;

\\\frac{y}{x}= \log\frac{kx}{(x-y)^{2}}\\ \frac{kx}{(x-y)^{2}}=e^{y/x}\\ (x-y)^{2}=kxe^{-y/x}
This is the required solution of given diff. equation

Question:2 Show that the given differential equation is homogeneousand solve each of them. y' = \frac{x+y}{x}

Answer:

the above differential eq can be written as,

\frac{dy}{dx} = F(x,y)=\frac{x+y}{x} ............................(i)

Now, F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x} = \lambda ^{0}F(x,y)
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx}= \frac{x+vx}{x} = 1+v
\\x\frac{dv}{dx}= 1\\ dv = \frac{dx}{x}
Integrating on both sides, we get; (and substitute the value of v =\frac{y}{x} )

\\v =\log x+C\\ \frac{y}{x}=\log x+C\\ y = x\log x +Cx
this is the required solution

Question:3 Show that the given differential equation is homogeneous and solve each of them.

(x-y)dy - (x+y)dx = 0

Answer:

The given differential eq can be written as;

\frac{dy}{dx}=\frac{x+y}{x-y} = F(x,y)(let\ say) ....................................(i)

F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}= \lambda ^{0}F(x,y)
Hence it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v+x\frac{dv}{dx}= \frac{1+v}{1-v}\\ x\frac{dv}{dx} = \frac{1+v}{1-v}-v =\frac{1+v^{2}}{1-v}

\frac{1-v}{1+v^{2}}dv = (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}
Integrating on both sides, we get;

\tan^{-1}v-1/2 \log(1+v^{2})=\log x+C
again substitute the value of v=y/x
\\\tan^{-1}(y/x)-1/2 \log(1+(y/x)^{2})=\log x+C\\ \tan^{-1}(y/x)-1/2 [\log(x^{2}+y^{2})-\log x^{2}]=\log x+C\\ tan^{-1}(y/x) = 1/2[\log (x^{2}+y^{2})]+C

This is the required solution.

Question:4 Show that the given differential equation is homogeneous and solve each of them.

(x^2 - y^2)dx + 2xydy = 0

Answer:

we can write it as;

\frac{dy}{dx}= -\frac{(x^{2}-y^{2})}{2xy} = F(x,y)\ (let\ say) ...................................(i)

F(\lambda x,\lambda y) = \frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)} = \lambda ^{0}.F(x,y)
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx} = \frac{ x^{2}-(vx)^{2}}{2x(vx)} =\frac{v^{2}-1}{2v}
\\x\frac{dv}{dx} =\frac{v^{2}+1}{2v}\\ \frac{2v}{1+v^{2}}dv=\frac{dx}{x}
integrating on both sides, we get

\log (1+v^{2})= -\log x +\log C = \log C/x
\\= 1+v^{2} = C/x\\ = x^2+y^{2}=Cx .............[ v =y/x ]
This is the required solution.

Question:5 Show that the given differential equation is homogeneous and solve it.

x^2\frac{dy}{dx} = x^2 - 2y^2 +xy

Answer:

\frac{dy}{dx}= \frac{x^{2}-2y^{2}+xy}{x^{2}} = F(x,y)\ (let\ say)

F(\lambda x,\lambda y)= \frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda .\lambda )xy}{(\lambda x)^{2}} = \lambda ^{0}.F(x,y) ............(i)
Hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v+x\frac{dv}{dx}= 1-2v^{2}+v\\ x\frac{dv}{dx} = 1-2v^{2}\\ \frac{dv}{1-2v^{2}}=\frac{dx}{x}

1/2[\frac{dv}{(1/\sqrt{2})^{2}-v^{2}}] = \frac{dx}{x}

On integrating both sides, we get;

\frac{1}{2\sqrt{2}}\log (\frac{1/\sqrt{2}+v}{1/\sqrt{2}-v}) = \log x +C
after substituting the value of v= y/x

\frac{1}{2\sqrt{2}}\log (\frac{x+\sqrt{2}y}{x-\sqrt{2}y}) = \log \left | x \right | +C

This is the required solution

Question:6 Show that the given differential equation is homogeneous and solve it.

xdy - ydx = \sqrt{x^2 + y^2}dx

Answer:

\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} = F(x,y) .................................(i)

F(\mu x,\mu y)=\frac{\mu y+\sqrt{(\mu x)^{2}+(\mu y)^{2}}}{\mu x} =\mu^{0}.F(x,y)
henxe it is a homogeneous equation

Now, to solve substitute y = vx

Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx}= v+\sqrt{1+v^{2}}=\sqrt{1+v^{2}}

=\frac{dv}{\sqrt{1+v^{2}}} =\frac{dx}{x}

On integrating both sides,

\Rightarrow \log \left | v+\sqrt{1+v^{2}} \right | = \log \left | x \right |+\log C
Substitute the value of v=y/x , we get

\\\Rightarrow \log \left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right | = \log \left | Cx \right |\\ y+\sqrt{x^{2}+y^{2}} = Cx^{2}

Required solution

Question:7 Solve.

\left\{x\cos\left(\frac{y}{x} \right ) + y\sin\left(\frac{y}{x} \right ) \right \}ydx = \left\{y\sin\left(\frac{y}{x} \right ) - x\cos\left(\frac{y}{x} \right ) \right \}xdy

Answer:

\frac{dy}{dx} =\frac{x \cos(y/x)+y\sin(y/x)}{y\sin(y/x)-x\cos(y/x)}.\frac{y}{x} = F(x,y) ......................(i)
By looking at the equation we can directly say that it is a homogenous equation.

Now, to solve substitute y = vx

Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx} =\frac{v \cos v+v^{2}\sin v}{v\sin v-\cos v}\\ =x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v-\cos v}\\ =(\tan v-1/v)dv = \frac{2dx}{x}

integrating on both sides, we get

\\=\log(\frac{\sec v}{v})= \log (Cx^{2})\\=\sec v/v =Cx^{2}
substitute the value of v= y/x , we get

\\\sec(y/x) =Cxy \\ xy \cos (y/x) = k

Required solution

Question:8 Solve.

x\frac{dy}{dx} - y + x\sin\left(\frac{y}{x}\right ) = 0

Answer:

\frac{dy}{dx}=\frac{y-x \sin(y/x)}{x} = F(x,y) ...............................(i)

F(\mu x, \mu y)=\frac{\mu y-\mu x \sin(\mu y/\mu x)}{\mu x} = \mu^{0}.F(x,y)
it is a homogeneous equation

Now, to solve substitute y = vx

Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

v+x\frac{dv}{dx}= v- \sin v = -\sin v
\Rightarrow -\frac{dv}{\sin v} = -(cosec\ v)dv=\frac{dx}{x}

On integrating both sides we get;

\\\Rightarrow \log \left | cosec\ v-\cot v \right |=-\log x+ \log C\\ \Rightarrow cosec (y/x) - \cot (y/x) = C/x

= x[1-\cos (y/x)] = C \sin (y/x) Required solution

Question:9 Solve.

ydx + x\log\left(\frac{y}{x} \right ) -2xdy = 0

Answer:

\frac{dy}{dx}= \frac{y}{2x-x \log(y/x)} = F(x,y) ..................(i)

\frac{\mu y}{2\mu x-\mu x \log(\mu y/\mu x)} = F(\mu x,\mu y) = \mu^{0}.F(x,y)

hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}= \frac{v}{2-\log v}\\ =x\frac{dv}{dx} = \frac{v\log v-v}{2-\log v}\\ =[\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}
integrating on both sides, we get; ( substituting v =y/x)

\\\Rightarrow \log[\log(y/x)-1]-\log(y/x)=\log(Cx)\\\Rightarrow \frac{x}{y}[\log(y/x)-1]=Cx\\ \Rightarrow \log (y/x)-1=Cy

This is the required solution of the given differential eq

Question:10 Solve.

\left(1 + e^{\frac{x}{y}} \right )dx + e^\frac{x}{y}\left(1-\frac{x}{y}\right )dy = 0

Answer:

\frac{dx}{dy}=\frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} = F(x,y) .......................................(i)

= F(\mu x,\mu y)=\frac{-e^{\mu x/\mu y}(1-\mu x/\mu y)}{1+e^{\mu x/\mu y}} =\mu^{0}.F(x,y)
Hence it is a homogeneous equation.

Now, to solve substitute x = yv

Diff erentiating on both sides wrt x
\frac{dx}{dy}= v +y\frac{dv}{dy}

Substitute this value in equation (i)

\\=v+y\frac{dv}{dy} = \frac{-e^{v}(1-v)}{1+e^{v}} \\ =y\frac{dv}{dy} = -\frac{v+e^{v}}{1+e^{v}}\\ =\frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}

Integrating on both sides, we get;

\dpi{100} \log(v+e^{v})=-\log y+ \log c =\log (c/y)\\ =[\frac{x}{y}+e^{x/y}]= \frac{c}{y}\\\Rightarrow x+ye^{x/y}=c
This is the required solution of the diff equation.

Question:11 Solve for particular solution.

(x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1

Answer:

\frac{dy}{dx}=\frac{-(x-y)}{x+y} =F(x,y) ..........................(i)

We can clearly say that it is a homogeneous equation.

Now, to solve substitute y = vx

Diff erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\v+x\frac{dv}{dx}=\frac{v-1}{v+1}\\ \Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}

\frac{1+v}{1+v^{2}}dv = [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv=-\frac{dx}{x}

On integrating both sides

\\=\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k\\ =\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k\\ =\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k\\ =\log(x^{2}+y^{2})+2\tan^{-1}(y/x) = 2k ......................(ii)

Now, y=1 and x= 1


\\=\log 2 +2\tan^{-1}1=2k\\ =\pi/2+\log 2 = 2k\\

After substituting the value of 2k in eq. (ii)

\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2

This is the required solution.

Question:12 Solve for particular solution.

x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1

Answer:

\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y) ...............................(i)

F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i), we get

\\=v+\frac{xdv}{dx}= -v- v^{2}\\ =\frac{xdv}{dx}=-v(v+2)\\ =\frac{dv}{v+2}=-\frac{dx}{x}\\ =1/2[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}

Integrating on both sides, we get;

\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\ =\frac{v}{v+2}=(C/x)^{2}

replace the value of v=y/x

\frac{x^{2}y}{y+2x}=C^{2} .............................(ii)

Now y =1 and x = 1

C = 1/\sqrt{3}
therefore,

\frac{x^{2}y}{y+2x}=1/3

Required solution

Question:13 Solve for particular solution.

\left [x\sin^2\left(\frac{y}{x} \right ) - y \right ]dx + xdy = 0;\ y =\frac{\pi}{4}\ when \ x = 1

Answer:

\frac{dy}{dx}=\frac{-[x\sin^{2}(y/x)-y]}{x} = F(x,y) ..................(i)

F(\mu x,\mu y)=\frac{-[\mu x\sin^{2}(\mu y/\mu x)-\mu y]}{\mu x}=\mu ^{0}.F(x,y)

Hence it is a homogeneous eq

Now, to solve substitute y = vx

Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

on integrating both sides, we get;

\\-\cot v =\log\left | x \right | -C\\ =\cot v = \log\left | x \right |+\log C

On substituting v =y/x

=\cot (y/x) = \log\left | Cx \right | ............................(ii)

Now, y = \pi/4\ @ x=1

\\\cot (\pi/4) = \log C \\ =C=e^{1}

put this value of C in eq (ii)

\cot (y/x)=\log\left | ex \right |

Required solution.

Question:14 Solve for particular solution.

\frac{dy}{dx} - \frac{y}{x} + \textup{cosec}\left (\frac{y}{x} \right ) = 0;\ y = 0 \ \textup{when}\ x = 1

Answer:

\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y) ....................................(i)

the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}=v- cosec\ v\\ =x\frac{dv}{dx} = -cosec\ v\\ =-\frac{dv}{cosec\ v}= \frac{dx}{x}\\ =-\sin v dv = \frac{dx}{x}

on integrating both sides, we get;

\\=cos\ v = \log x +\log C =\log Cx\\ =\cos(y/x)= \log Cx .................................(ii)

now y = 0 and x =1 , we get

C =e^{1}

put the value of C in eq 2

\cos(y/x)=\log \left | ex \right |

Question:15 Solve for particular solution.

2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1

Answer:

The above eq can be written as;

\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)
By looking, we can say that it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}

Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}= \frac{2v+v^{2}}{2}\\ =x\frac{dv}{dx} = v^{2}/2\\ = \frac{2dv}{v^{2}}=\frac{dx}{x}

integrating on both sides, we get;

\\=-2/v=\log \left | x \right |+C\\ =-\frac{2x}{y}=\log \left | x \right |+C .............................(ii)

Now, y = 2 and x =1, we get

C =-1
put this value in equation(ii)

\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}

Question:16 A homogeneous differential equation of the from \frac{dx}{dy}= h\left(\frac{x}{y} \right ) can be solved by making the substitution.

(A) y = vx

(B) v = yx

(C) x = vy

(D) x =v

Answer:

\frac{dx}{dy}= h\left(\frac{x}{y} \right )
for solving this type of equation put x/y = v
x = vy

option C is correct

Question:17 Which of the following is a homogeneous differential equation?

(A) (4x + 6x +5)dy - (3y + 2x +4)dx = 0

(B) (xy)dx - (x^3 + y^3)dy = 0

(C) (x^3 +2y^2)dx + 2xydy =0

(D) y^2dx + (x^2 -xy -y^2)dy = 0

Answer:

Option D is the right answer.

y^2dx + (x^2 -xy -y^2)dy = 0
\frac{dy}{dx}=\frac{y^{2}}{x^{2}-xy-y^{2}} = F(x,y)
we can take out lambda as a common factor and it can be cancelled out

More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

To practice the concepts given in topic 9.5.2, 4 example questions are given prior to the exercise 9.5 Class 12 Maths. There are 17 questions in the Class 12 Maths chapter 9 exercise 9.5 and 2 questions of this have 4 choices. As mentioned in the first para certain questions in the Class 12th Maths chapter 6 exercise 9.5 are to find the general solutions and some questions are to find the particular solutions of homogeneous differential equations.

Also Read| Differential Equations Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

  • Utilising the exercise 9.5 Class 12 Maths students can master the concepts of identifying the homogeneous equations and finding the solutions to these equations.
  • Not only for CBSE board exams but also for various state board exams and Engineering Entrance Exams, the NCERT solutions for Class 12 Maths chapter 9 exercise 9.5 can be used.
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Key Features Of NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 9.5 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 9.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 9.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 9.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 9.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 9.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Questions (FAQs)

1. What number of multiple-choice questions are present in the exercise 9.5 Class 12 Maths?

Two objective type questions with 4 choices are given in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.5.

2. What is the first step in solving the problems given in the Class 12 Maths chapter 9 exercise 9.5?

The first step is to verify the given differential equation is homogeneous or not.

3. How many examples are solved in topic 9.5.2?

4 examples are given.

4. Whether the concept of integration is required to solve the problems of Class 12th Maths chapter 6 exercise 9.5?

Yes, the concepts of basic integrals studied in the NCERT Class 12 Maths chapter 7 is required to solve exercise 9.5

5. What is the topic discussed before the homogeneous differrential equation?

The differential equations of  variable separable form.

6. The number of exercises coming under 9.5 are……….

Three exercises are coming under the topic .5

7. What is the topic discussed after exercise 9.5?

The idea of a linear differential equation is discussed after exercise 9.5 Class 12 Maths.

8. How many questions are present in exercise 9.5?

Five

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i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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kumardurgesh1802 10 July,2024
I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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satyamercedes0626 23 July,2024
I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

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Abhishek 12 July,2024

Popular CBSE Class 12th Questions

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
Read More

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

 
Read More

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
Read More

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
Read More

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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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