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

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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

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.
$frac { dy }{ dx } +y=1(yneq 1)$
Solution:
$frac { dy }{ dx } +y=1Rightarrow int { frac { dy }{ y-1 } } =-int { dx }$
$Rightarrow log(y-1)=-x+cRightarrow y=1+{ e }^{ -x }.{ e }^{ c }$
$Hencequad y=1+{ Ae }^{ -x }$
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

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

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:
$because quad yquad logyquad dx=xquad dyRightarrow frac { dy }{ yquad logy } =frac { dx }{ x }$
integrating we get

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

Ex 9.4 Class 12 Maths Question 9.
solve the following
$frac { dy }{ dx } ={ sin }^{ -1 }x$
Solution:
$frac { dy }{ dx } ={ sin }^{ -1 }xRightarrow int { dy } =int { { sin }^{ -1 }xdx }$
integrating both sides we get

Ex 9.4 Class 12 Maths Question 10.
${ e }^{ x }tanyquad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0$
Solution:
${ e }^{ x }tanyquad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0$
we can write in another form

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.
$left( { x }^{ 3 }+{ x }^{ 2 }+x+1 right) frac { dy }{ dx } ={ 2x }^{ 2 }+x;y=1,whenquad x=0$
Solution:
here
$dy=frac { { 2x }^{ 2 }+x }{ left( { x }^{ 3 }+{ x }^{ 2 }+x+1 right) } dx$
integrating we get

Ex 9.4 Class 12 Maths Question 12.
$xleft( { x }^{ 2 }-1 right) frac { dy }{ dx } =1,y=0quad whenquad x=2$
Solution:
$xleft( { x }^{ 2 }-1 right) frac { dy }{ dx } =1,y=0quad whenquad x=2$
$Rightarrow int { dy } =int { frac { dy }{ x(x+1)(x-1) } }$

Ex 9.4 Class 12 Maths Question 13.
$cosleft( frac { dy }{ dx } right) =a,(aepsilon R),y=1quad whenquad x=0$
Solution:
$cosleft( frac { dy }{ dx } right) =aquad therefore frac { dy }{ dx } ={ cos }^{ -1 }a$

Ex 9.4 Class 12 Maths Question 14.
$frac { dy }{ dx } =ytanx,y=1quad whenquad x=0$
Solution:
$frac { dy }{ dx } =ytanxRightarrow int { frac { dy }{ y } } =int { tanxquad dx }$
=> 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:
${ y }^{ I }={ e }^{ x }sinx$
$Rightarrow dy={ e }^{ x }sinxquad dx$

Ex 9.4 Class 12 Maths Question 16.
For the differential equation $xyfrac { dy }{ dx } =(x+2)(y+2)$ find the solution curve passing through the point (1,-1)
Solution:
The differential equation is $xyfrac { dy }{ dx } =(x+2)(y+2)$
or xydy=(x + 2)(y+2)dx

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 $yfrac { dy }{ dx } =x$
$Rightarrow int { ydy } =int { xdx } Rightarrow frac { { y }^{ 2 } }{ 2 } =frac { { x }^{ 2 } }{ 2 } +c$
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)

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.

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

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%

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

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) $frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }Rightarrow int { { e }^{ -y }dy } =int { { e }^{ x }dx }$
$Rightarrow { e }^{ -y }={ e }^{ x }+kRightarrow { e }^{ x }+{ e }^{ -y }=c$

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

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