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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4.

Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 9
Chapter Name Differential Equations
Exercise Ex 9.4
Number of Questions Solved 23
Category NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

For each of the following D.E in Q. 1 to 10 find the general solution:

Ex 9.4 Class 12 Maths Question 1.
\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx }
Solution:
\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx }
\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx } =\frac { { 2sin }^{ 2 }\left( \frac { x }{ 2 } \right) }{ { 2cos }^{ 2 }\left( \frac { x }{ 2 } \right) } ={ tan }^{ 2 }\left( \frac { x }{ 2 } \right)
integrating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Ex 9.4 Class 12 Maths Question 2.
\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } (-2<y<2)
Solution:
\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } \Rightarrow \int { \frac { dy }{ \sqrt { { 4-y }^{ 2 } } } } =\int { dx }
\Rightarrow { sin }^{ -1 }\frac { y }{ 2 } =x+C
\Rightarrow y=2sin(x+C)

Ex 9.4 Class 12 Maths Question 3.
\frac { dy }{ dx } +y=1(y\neq 1)
Solution:
\frac { dy }{ dx } +y=1\Rightarrow \int { \frac { dy }{ y-1 } } =-\int { dx }
\Rightarrow log(y-1)=-x+c\Rightarrow y=1+{ e }^{ -x }.{ e }^{ c }
Hence\quad y=1+{ Ae }^{ -x }
which is required solution

Ex 9.4 Class 12 Maths Question 4.
sec² x tany dx+sec² y tanx dy = 0
Solution:
we have
sec² x tany dx+sec² y tanx dy = 0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Ex 9.4 Class 12 Maths Question 5.
\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0
Solution:
we have
\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0
Integrating on both sides
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 5

Ex 9.4 Class 12 Maths Question 6.
\frac { dy }{ dx } =\left( { 1+x }^{ 2 } \right) \left( { 1+y }^{ 2 } \right)
Solution:
\frac { dy }{ { 1+y }^{ 2 } } =\left( { 1+x }^{ 2 } \right) dx
integrating on both side we get
{ tan }^{ -1 }y={ x+\frac { 1 }{ 3 } }x^{ 3 }+c
which is required solution

Ex 9.4 Class 12 Maths Question 7.
y logy dx – x dy = 0
Solution:
\because \quad y\quad logy\quad dx=x\quad dy\Rightarrow \frac { dy }{ y\quad logy } =\frac { dx }{ x }
integrating we get
tiwari academy class 12 maths Chapter 9 Differential Equations 7

Ex 9.4 Class 12 Maths Question 8.
{ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }
Solution:
{ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }\Rightarrow \int { { y }^{ -5 }dy } =-\int { { x }^{ -5 }dx }
\Rightarrow -\frac { 1 }{ { y }^{ 4 } } =\frac { 1 }{ { x }^{ 4 } } +4c\Rightarrow { x }^{ -4 }+{ y }^{ -4 }=k

Ex 9.4 Class 12 Maths Question 9.
solve the following
\frac { dy }{ dx } ={ sin }^{ -1 }x
Solution:
\frac { dy }{ dx } ={ sin }^{ -1 }x\Rightarrow \int { dy } =\int { { sin }^{ -1 }xdx }
integrating both sides we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9

Ex 9.4 Class 12 Maths Question 10.
{ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0
Solution:
{ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0
we can write in another form
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10

Find a particular solution satisfying the given condition for the following differential equation in Q.11 to 14.

Ex 9.4 Class 12 Maths Question 11.
\left( { x }^{ 3 }+{ x }^{ 2 }+x+1 \right) \frac { dy }{ dx } ={ 2x }^{ 2 }+x;y=1,when\quad x=0
Solution:
here
dy=\frac { { 2x }^{ 2 }+x }{ \left( { x }^{ 3 }+{ x }^{ 2 }+x+1 \right) } dx
integrating we get
tiwari academy class 12 maths Chapter 9 Differential Equations 11

Ex 9.4 Class 12 Maths Question 12.
x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2
Solution:
x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2
\Rightarrow \int { dy } =\int { \frac { dy }{ x(x+1)(x-1) } }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12.1

Ex 9.4 Class 12 Maths Question 13.
cos\left( \frac { dy }{ dx } \right) =a,(a\epsilon R),y=1\quad when\quad x=0
Solution:
cos\left( \frac { dy }{ dx } \right) =a\quad \therefore \frac { dy }{ dx } ={ cos }^{ -1 }a
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Ex 9.4 Class 12 Maths Question 14.
\frac { dy }{ dx } =ytanx,y=1\quad when\quad x=0
Solution:
\frac { dy }{ dx } =ytanx\Rightarrow \int { \frac { dy }{ y } } =\int { tanx\quad dx }
=> logy = logsecx + C
When x = 0, y = 1
=> log1 = log sec0 + C => 0 = log1 + C
=> C = 0
∴ logy = log sec x
=> y = sec x.

Ex 9.4 Class 12 Maths Question 15.
Find the equation of the curve passing through the point (0,0) and whose differential equation { y }^{ I }={ e }^{ x }sinx
Solution:
{ y }^{ I }={ e }^{ x }sinx
\Rightarrow dy={ e }^{ x }sinx\quad dx
tiwari academy class 12 maths Chapter 9 Differential Equations 15

Ex 9.4 Class 12 Maths Question 16.
For the differential equation xy\frac { dy }{ dx } =(x+2)(y+2) find the solution curve passing through the point (1,-1)
Solution:
The differential equation isxy\frac { dy }{ dx } =(x+2)(y+2)
or xydy=(x + 2)(y+2)dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 16

Ex 9.4 Class 12 Maths Question 17.
Find the equation of a curve passing through the point (0, -2) given that at any point (pc, 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
Solution:
According to the question y\frac { dy }{ dx } =x
\Rightarrow \int { ydy } =\int { xdx } \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c
0, – 2) lies on it.c = 2
∴ Equation of the curve is : x² – y² + 4 = 0.

Ex 9.4 Class 12 Maths 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).
Solution:
Slope of the tangent to the curve = \frac { dy }{ dx }
slope of the line joining (x, y) and (- 4, – 3)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 18
tiwari academy class 12 maths Chapter 9 Differential Equations 18.1

Ex 9.4 Class 12 Maths Question 19.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution:
Let v be volume of the balloon.
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 19

Ex 9.4 Class 12 Maths Question 20.
In a bank principal increases at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years
Solution:
Let P be the principal at any time t.
According to the problem
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 20

Ex 9.4 Class 12 Maths Question 21.
In a bank principal increases 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
Solution:
Let p be the principal Rate of interest is 5%
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 21

Ex 9.4 Class 12 Maths 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
Solution:
Let y denote the number of bacteria at any instant t • then according to the question
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 22

Ex 9.4 Class 12 Maths Question 23.
The general solution of a differential equation \frac { dy }{ dx } ={ e }^{ x+y } is
(a) { e }^{ x }+{ e }^{ -y }=c
(b) { e }^{ x }+{ e }^{ y }=c
(c) { e }^{ -x }+{ e }^{ y }=c
(d) { e }^{ -x }+{ e }^{ -y }=c
Solution:
(a) \frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }\Rightarrow \int { { e }^{ -y }dy } =\int { { e }^{ x }dx }
\Rightarrow { e }^{ -y }={ e }^{ x }+k\Rightarrow { e }^{ x }+{ e }^{ -y }=c

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