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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
  7. Also see-
  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.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

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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quartely question paper for mathematics cbse

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Jefferson 25 September,2024
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  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

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 assist you.

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

Read Complete Answer
kumardurgesh1802 10 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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October 11, 2024 - 09:10 PM IST :
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