Differential Equations Class 12 Mathematics Extra Questions

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2024-25, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Install Now

Differential Equations Class 12 Mathematics Extra Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 9 Differential Equations Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 9 Maths Extra Questions

Download as PDF

Chapter 9 Mathematics Class 12 Important Questions

Chapter 9 Differential Equations

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 (loge2 = 0.6931).
1. 9.93%
2. 7.93%
3. 6.93%
4. 8.93%
General solution of{tex}co{s^2}x\frac{{dy}}{{dx}} + y = \;\tan x\;\left( {0 \leqslant x < \frac{\pi }{2}} \right){/tex} is
1. {tex}y = \left( {\tan x – 1} \right) + C{e^{ – \tan x}}{/tex}
2. {tex}y = \left( {\tan x + 1} \right) + C{e^{ – \tan x}}{/tex}
3. {tex}y = \left( {\tan x + 1} \right) – C{e^{ – \tan x}}{/tex}
4. {tex}y = \left( {\tan x – 1} \right) – C{e^{ – \tan x}}{/tex}
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
1. 3
2. 2
3. 1
4. 4
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 {tex}\left( {{e^{0.5}} = {\text{ }}1.648} \right).{/tex}
1. Rs 1848
2. Rs 1648
3. Rs 1748
4. Rs 1948
What is the order of differential equation : {tex}\frac{{{d^3}y}}{{d{x^3}}} + \frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = {e^x}{/tex}.
1. 2
2. 3
3. 1
4. 0
F(x, y) = {tex}\frac{{\sqrt {{x^2} + {y^2}} + y}}{x}{/tex} is a homogeneous function of degree ________.
The degree of the differential equation {tex}\sqrt{1+\left(\frac{dy}{dx}\right)^2}=x{/tex} is ________.
The order of the differential equation of all circles of given radius a is ________.
Verify that the function is a solution of the corresponding differential equation {tex}y = x\sin x;\,\,x{y^,} = y + x\sqrt {{x^2} – {y^2}} {/tex}.
Find order and degree. {tex}\frac{{{d^4}y}}{{d{x^2}}} + \sin \left( {y”’} \right) = 0{/tex}.
Write the solution of the differential equation {tex}\frac { d y } { d x } = 2 ^ { – y }{/tex}.
Verify that the given function (explicit) is a solution of the corresponding differential equation: y = x² + 2x + C : y’ – 2x – 2 = 0.
Find the differential equation of all non-horizontal lines in a plane.
Verify that the function is a solution of the corresponding differential equation
{tex}y = \sqrt {1 + {x^2}} ;{y’} = \frac{{xy}}{{1 + {x^2}}}{/tex}.
Solve the following differential equation.
{tex} \left( y + 3 x ^ { 2 } \right) \frac { d x } { d y } = x{/tex}
Solve the differential equation (1 + y²) tan^-1x dx + 2y (1 + x²) dy = 0.
Find the particular solution of the differential equation (1 + e^2x)dy + (1 + y²)e^x dx = 0, given that y = 1, when x = 0.
Solve {tex}\left( {1 + {e^{\frac{x}{y}}}} \right)dx + {e^{\frac{x}{y}}}\left( {1 – \frac{x}{y}} \right)dy = 0{/tex}.

Chapter 9 Differential Equations

Solution

1. 6.93%
  Explanation: Let P be the principal at any time t. then,
  {tex}\frac{{dP}}{{dt}} = \frac{{rP}}{{100}} \Rightarrow \frac{{dP}}{{dt}} = \frac{P}{{100}}{/tex}
  {tex} \Rightarrow \int {\frac{1}{P}dP = \int {\frac{r}{{100}}dt} } {/tex}
  {tex} \Rightarrow \log P = \frac{r}{{100}}t + \log c{/tex}
  {tex} \Rightarrow \log \frac{P}{c} = \frac{r}{{100}}t{/tex}
  {tex}\Rightarrow P = c{e^{\frac{r}{{100}}}}{/tex}
  When P = 100 and t = 0., then, c = 100, therefore, we have:
  {tex}\Rightarrow P=100 \ {{e}^{{}^{r}\!\!\diagup\!\!{}_{100}\;}}{/tex}
  Now, let t = T, when P = 100., then;
  {tex}\Rightarrow 200 = 100{e^{\frac{T}{{100}}}}{/tex}
  {tex} \Rightarrow {e^{\frac{T}{{100}}}} = 2{/tex}
  {tex} \Rightarrow T = 100\log 2{/tex} = 100(0.6931) = 6.93%
1. {tex}y = \left( {\tan x – 1} \right) + C{e^{ – \tan x}}{/tex}
  Explanation: {tex}\frac{{dy}}{{dx}} + {\sec ^2}x.y = \tan x.{\sec ^2}x \Rightarrow P = {\sec ^2}x,Q = \tan x.{\sec ^2}x{/tex}
  {tex} \Rightarrow I.F. = {e^{\int_{}^{} {{{\sec }^2}xdx} }} = {e^{\tan x}}{/tex}
  {tex}\Rightarrow y.{e^{\tan x}} = \int_{}^{} {\tan x{{\sec }^2}x{e^{\tan x}}dx \Rightarrow y.{e^{\tan x}} = (\tan x – 1){e^{\tan x}} + C} {/tex}
  {tex} \Rightarrow y = (\tan x – 1) + C{e^{ – \tan x}}{/tex}
1. 4
  Explanation: 4, because the no. of arbitrary constants is equal to order of the differential equation.
1. Rs 1648
  Explanation: Here P is the principal at time t
  {tex}\frac{{dP}}{{dt}} = \frac{{5P}}{{100}} \Rightarrow \frac{{dP}}{{dt}} = \frac{P}{{20}}{/tex}
  {tex} \Rightarrow \int_{}^{} {\frac{1}{P}dP = \int_{}^{} {\frac{1}{{20}}dt} } {/tex}
  {tex}\Rightarrow \log P = \frac{1}{{20}}t + \log c{/tex}
  {tex} \Rightarrow \log \frac{P}{c} = \frac{1}{{20}}t{/tex}
  {tex}\Rightarrow P =c {e^{\frac{1}{{100}}}}{/tex}
  When P = 1000 and t = 0 ., then ,
  c = 1000, therefore, we have :
  {tex} \Rightarrow P = 1000{e^{\frac{T}{{100}}}}{/tex}
  {tex}\Rightarrow A = 1000{e^{\frac{5}{{10}}}}{/tex}
  {tex}\Rightarrow {e^{\frac{5}{{10}}}} = A{/tex}
  {tex} \Rightarrow A = 1000\log 0.5{/tex}
  = 1000(1.648)
  = 1648
1. 3
  Explanation: Order = 3. Since the third derivative is the highest derivative present in the equation. i.e.{tex}\frac{{{d^3}y}}{{d{x^3}}}{/tex}
Zero
not defined
2
{tex}y = x.\sin x{/tex}…(1)
{tex}{y^,} = x.\cos x + \sin x.1{/tex}
{tex} \Rightarrow x{y^,} = {x^2}\cos x + x.\sin x{/tex}
{tex}x{y^,} = {x^2}\sqrt {1 – {{\sin }^2}x} + x.\sin x{/tex}
{tex}x{y^,} = {x^2}\sqrt {1 – {{\left( {\frac{y}{x}} \right)}^2}} + x.\sin x{/tex} {tex}\left[ {\because \frac{y}{x} = \sin x} \right.]{/tex}
{tex}x{y^,} = {x^2}\frac{{\sqrt {{x^2} – {y^2}} }}{x} + x.\sin x{/tex}
{tex}x{y^,} = x\sqrt {{x^2} – {y^2}} + y{/tex}
Hence proved.
order = 4 ,degree = not defined
Given differential equation is
{tex}\frac { d y } { d x } = 2 ^ { -y }{/tex}
on separating the variables, we get
2^ydy = dx
On integrating both sides, we get
{tex}\int 2 ^ { y } d y = \int d x{/tex}
{tex}\Rightarrow \quad \frac { 2 ^ { y } } { \log 2 } = x + C_1{/tex}
{tex}\Rightarrow{/tex} 2^y = x log 2 + C₁ log 2
{tex}\therefore{/tex} 2^y = x log 2 + C, where C = C₁ log 2
Given: y = x² + 2x + C …(i)
To prove: y is a solution of the differential equation y’ – 2x – 2 = 0 …(ii)
Proof:From, eq. (i),
y’ = 2x + 2
L.H.S. of eq. (ii),
= y’ – 2x – 2
= (2x + 2) – 2x – 2
= 2x + 2 – 2x – 2 = 0 = R.H.S.
Hence, y given by eq. (i) is a solution of y’ – 2x – 2 = 0.
The general equation of all non-horizontal lines in a plane is ax + by = c, where {tex}a \ne 0{/tex}.
differentiating both sides w.r.t. y on both sides,we get
{tex}a\frac{{dy}}{{dx}} + b = 0{/tex}
Again, differentiating both sides w.r.t. y, we get
{tex}a\frac{{{d^2}x}}{{d{y^2}}} = 0 \Rightarrow \frac{{{d^2}x}}{{d{y^2}}} = 0.{/tex}
{tex}y = \sqrt {1 + {x^2}} {/tex} ……(i)
{tex}{y’} = \frac{1}{{2\sqrt {1 + {x^2}} }}.2x{/tex} ……(ii)
{tex}(ii) \div (i){/tex},we get,
{tex}\Rightarrow\frac{{{y’}}}{y} = \frac{{\frac{x}{{\sqrt {1 + {x^2}} }}}}{{\sqrt {1 + {x^2}} }}{/tex}
{tex}\Rightarrow\frac{{{y’}}}{y} = \frac{x}{{1 + {x^2}}}{/tex}
{tex}{y’} = \frac{{xy}}{{1 + {x^2}}}{/tex}
Hence given value of y is the solution of given differential equation.
According to the question,we have to solve the differential equation ,
{tex} \left( y + 3 x ^ { 2 } \right) \frac { d x } { d y } = x \Rightarrow \frac { d y } { d x } = \frac { y } { x } + 3 x{/tex}
{tex} \Rightarrow \quad \frac { d y } { d x } – \frac { y } { x } = 3 x{/tex}
which is a linear differential equation of the form
{tex} \frac { d y } { d x } + P y = Q{/tex}.
Here, {tex} P = \frac { – 1 } { x }{/tex}and Q = 3x
{tex} \therefore \quad \mathrm { IF } = e ^ { \int P d x } = e ^ { \int – \frac { 1 } { x } d x } = e ^ { – \log | x | } = e ^ { \log x ^ { – 1 } } = x ^ { – 1 }{/tex}
{tex} \Rightarrow \quad \mathrm { IF } = x ^ { – 1 } = \frac { 1 } { x }{/tex}
The solution of linear differential equation is given by
{tex} y \times I F = \int ( Q \times I F ) d x + C{/tex}
{tex} \Rightarrow \quad y \times \frac { 1 } { x } = \int \left( 3 x \times \frac { 1 } { x } \right) d x+C{/tex}
{tex} \Rightarrow \quad \frac { y } { x } = \int 3 d x+C \Rightarrow \frac { y } { x } = 3 x + C{/tex}
{tex} \therefore \quad y = 3 x ^ { 2 } + C x{/tex}
which is the required solution.
Given differential equation is
(1 + y²) tan^-1x dx + 2y (1 + x²) dy = 0
{tex}\Rightarrow \left( {1 + {y^2}} \right){\tan ^{ – 1}}xdx = – 2y\left( {1 + {x^2}} \right)dy{/tex}
{tex}\Rightarrow \frac{{{{\tan }^{ – 1}}xdx}}{{1 + {x^2}}} = – \frac{{2y}}{{1 + {y^2}}}dy{/tex}
On integrating both sides, we get
{tex}\int {\frac{{{{\tan }^{ – 1}}x}}{{1 + {x^2}}}dx = – \int {\frac{{2y}}{{1 + {y^2}}}dy} } {/tex}
Put {tex}{\tan ^{ – 1}}x = t\;in\,\,LHS,{/tex} we get
{tex}\frac{1}{{1 + {x^2}}}dx = dt{/tex}
and put 1 + y² = u in RHS, we get
2ydy = du
{tex} \Rightarrow \int {tdt = – \int {\frac{1}{u} \Rightarrow \frac{{{t^2}}}{2} = – \log u + C} } {/tex}
{tex}\Rightarrow \frac{1}{2}{\left( {{{\tan }^{ – 1}}x} \right)^2} = – \log \left( {1 + {y^2}} \right) + C{/tex}
{tex}\Rightarrow \frac{1}{2}{\left( {{{\tan }^{ – 1}}x} \right)^2} + \log \left( {1 + {y^2}} \right) = C{/tex}
Given differential equation is,
(1 + e^2x)dy + (1 + y²)e^x dx = 0
Above equation may be written as
{tex}\frac { d y } { 1 + y ^ { 2 } } = \frac { – e ^ { x } } { 1 + e ^ { 2 x } } d x{/tex}
On integrating both sides, we get
{tex}\int \frac { d y } { 1 + y ^ { 2 } } = – \int \frac { e ^ { x } } { 1 + e ^ { 2 x } } d x{/tex}
On putting e^x = t {tex}\Rightarrow{/tex} e^x dx = dt in RHS, we get
{tex}\tan ^ { – 1 } y = – \int \frac { 1 } { 1 + t ^ { 2 } } d t{/tex}
{tex}\Rightarrow \quad \tan ^ { – 1 } y = – \tan ^ { – 1 } t + C{/tex}
{tex}\Rightarrow \quad \tan ^ { – 1 } y = – \tan ^ { – 1 } \left( e ^ { x } \right) + C{/tex} …(i) [put t = e^x] Also, given that y = 1, when x = 0.
On putting above values in Eq. (i), we get
tan^-11 = -tan^-1(e⁰) + C
{tex}\Rightarrow \quad \tan ^ { – 1 }1 = – \tan ^ { – 1 } 1 + C \quad \left[ \because e ^ { 0 } = 1 \right]{/tex}
{tex}\Rightarrow \quad 2 \tan ^ { – 1 } 1 = C{/tex}
{tex}\Rightarrow \quad 2 \tan ^ { – 1 } \left( \tan \frac { \pi } { 4 } \right) = C{/tex}
{tex}\Rightarrow \quad C = 2 \times \frac { \pi } { 4 } = \frac { \pi } { 2 }{/tex}
On putting {tex}C = \frac { \pi } { 2 }{/tex} in Eq. (i), we get
{tex}\tan ^ { – 1 } y = – \tan ^ { – 1 } e ^ { x } + \frac { \pi } { 2 }{/tex}
{tex}\Rightarrow \quad y = \tan \left[ \frac { \pi } { 2 } – \tan ^ { – 1 } \left( e ^ { x } \right) \right] = \cot \left[ \tan ^ { – 1 } \left( e ^ { x } \right) \right]{/tex}
{tex}= \cot \left[ \cot ^ { – 1 } \left( \frac { 1 } { e ^ { x } } \right) \right] \left[ \because \tan ^ { – 1 } x = \cot ^ { – 1 } \frac { 1 } { x } \right]{/tex}
{tex}\therefore \quad y = \frac { 1 } { e ^ { x } }{/tex}
which is the required solution.
{tex}\left( {1 + {e^{\frac{x}{y}}}} \right)dx + {e^{\frac{x}{y}}}\left( {1 – \frac{x}{y}} \right)dy = 0{/tex}
{tex}\Rightarrow\frac{{dx}}{{dy}} = – \frac{{{e^{x/y}}\left( {1 – \frac{x}{y}} \right)}}{{1 + {e^{x/y}}}}{/tex}
{tex}\Rightarrow\frac{{dx}}{{dy}} = \frac{{{e^{x/y}}\left( {\frac{x}{y} – 1} \right)}}{{1 + {e^{x/y}}}}{/tex}……..(1)
Let x = vy, then,
{tex}\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}{/tex}
Put {tex}\frac{{dx}}{{dy}}{/tex} in eq (1),we get,
{tex}v + y \frac{{dv}}{{dy}} = \frac{{{e^v}(v – 1)}}{{{e^v} + 1}}{/tex}
{tex}\Rightarrow y\frac{{dv}}{{dy}} = \frac{{v{e^v} – {e^v}}}{{{e^v} + 1}} – v{/tex}
{tex}\Rightarrow y\frac{{dv}}{{dy}} = \frac{{v{e^v} – {e^v} – v{e^v} – v}}{{{e^v} + 1}}{/tex}
{tex} \Rightarrow- \int {\frac{{dy}}{y}} = \int {\frac{{{e^v} + 1}}{{v + {e^v}}}dv} {/tex}
{tex}\Rightarrow\log ({e^v} + v) = – \log (y) + c{/tex}
{tex}\Rightarrow\log (({e^v} + v).y) = c{/tex}
{tex}\Rightarrow({e^v} + v)y = {e^c}{/tex}
{tex}\Rightarrow({e^v} + v)y = A{/tex} [Putting e^c = A] {tex}\Rightarrow\left( {{e^{x/y}} + \frac{x}{y}} \right)y = A{/tex}
{tex}\Rightarrow y{e^{x/y}} + x = A{/tex}

Chapter Wise Important Questions Class 12 Maths Part I and Part II

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Create Now

myCBSEguide

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes