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

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

NCERT TEXTBOOK EXERCISES

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:

$\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } \Rightarrow \int { \frac { dy }{ \sqrt { { 4-y }^{ 2 } } } } =\int { dx }$
â‡’ $\Rightarrow { sin }^{ -1 }\frac { y }{ 2 } =x+C$
â‡’ $\Rightarrow y=2sin(x+C)$

Ex 9.4 Class 12 Maths Question 3.
$\frac { dy }{ dx } +y=1(y\neq 1)$

Solution:

$\frac { dy }{ dx } +y=1\Rightarrow \int { \frac { dy }{ y-1 } } =-\int { dx }$
$\Rightarrow log(y-1)=-x+c\Rightarrow y=1+{ e }^{ -x }.{ e }^{ c }$
Hencey=1+Aeâˆ’x
which is required solution

Ex 9.4 Class 12 Maths Question 4.
secÂ² x tany dx+secÂ² y tanx dy = 0

Solution:

we have
secÂ² x tany dx+secÂ² 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 â€“ x dy = 0

Solution

Undefined control sequence \because
integrating we get

Ex 9.4 Class 12 Maths Question 8.
${ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }$

Solution:

${ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }\Rightarrow \int { { y }^{ -5 }dy } =-\int { { x }^{ -5 }dx }$
$\Rightarrow -\frac { 1 }{ { y }^{ 4 } } =\frac { 1 }{ { x }^{ 4 } } +4c\Rightarrow { x }^{ -4 }+{ y }^{ -4 }=k$

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

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

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

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:

$x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2$
â‡’ $\Rightarrow \int { dy } =\int { \frac { dy }{ x(x+1)(x-1) } }$

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:

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
âˆ´ 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:

${ y }^{ I }={ e }^{ x }sinx$
$\Rightarrow dy={ e }^{ x }sinx\quad dx$

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 is $xy\frac { dy }{ dx } =(x+2)(y+2)$
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 $y\frac { dy }{ dx } =x$
$\Rightarrow \int { ydy } =\int { xdx } \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c$
0, â€“ 2) lies on it.c = 2
âˆ´ Equation of the curve is : xÂ² â€“ yÂ² + 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 = dy/dx
slope of the line joining (x, y) and (- 4, â€“ 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 â€¢ 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) ex+eâˆ’y=c
(b) ex+ey=c
(c) eâˆ’x+ey=c
(d) eâˆ’x+eâˆ’y=c

Solution:

(a) $\frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }\Rightarrow \int { { e }^{ -y }dy } =\int { { e }^{ x }dx }$
$\Rightarrow { e }^{ -y }={ e }^{ x }+k\Rightarrow { e }^{ x }+{ e }^{ -y }=c$

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.4 PDF

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