NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 9
Chapter Name	Differential Equations
Exercise	Ex 9.6
Number of Questions Solved	19
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

Find the general solution of the following differential equations in Q.1 to 12

Question 1.
$\frac { dy }{ dx } +2y=sinx$
Solution:
Given equation is a linear differential equation of the form $\frac { dy }{ dx } +Py=Q$ ;
Here, P = 2, Q = sin x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Question 2.
$\frac { dy }{ dx } +3y={ e }^{ -2x }$
Solution:
$\frac { dy }{ dx } +3y={ e }^{ -2x }$
Here P = 3, $IF={ e }^{ \int { p.dx } }={ e }^{ 3x }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2
which is required equation

Question 3.
$\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }$
Solution:
$\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }$
$IF={ e }^{ \int { \frac { 1 }{ x } dx } }={ e }^{ logx }=x$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Question 4.
$\frac { dy }{ dx } +(secx)y=tanx\left( 0\le x<\frac { \pi }{ 2 } \right)$
Solution:
Here, P = secx, Q = tanx; $IF={ e }^{ \int { p.dx } }={ e }^{ \int { secx.dx } }$
$={ e }^{ log|secx+tanx| }$
= sec x + tan x
i.e., The solu. is y.× I.F. = ∫Q × I.F. dx + c
or y × (secx+tanx) = ∫tanx(secx+tanx)dx+c
Reqd. sol. is
∴ y(secx + tanx) = (secx + tanx)-x + c

Question 5.
${ cos }^{ 2 }x\frac { dy }{ dx } +y=tanx\left( 0\le x\le \frac { \pi }{ 2 } \right)$
Solution:
$\frac { dy }{ dx } +{ y\quad sec }^{ 2 }x={ sec }^{ 2 }x\quad tanx$
⇒ integrating factor = ${ e }^{ \int { { sec }^{ 2 }xdx } }={ e }^{ tanx }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 5

Question 6.
$x\frac { dy }{ dx } +2y={ x }^{ 2 }logx$
Solution:
$\frac { dy }{ dx } +\frac { 2 }{ x } y\quad =\quad x\quad logx$
Here P = $\frac { 2 }{ x }$ and Q = x logx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Question 7.
$xlogx\frac { dy }{ dx } +y=\frac { 2 }{ x } logx$
Solution:
$\frac { dy }{ dx } +\frac { 1 }{ xlogx } y=\frac { 2 }{ { x }^{ 2 } }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Question 8.
(1+x²)dy+2xy dx = cotx dx(x≠0)
Solution:
(1+x²)dy+2xy dx = cotx dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Question 9.
$x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)$
Solution:
$x\frac { dy }{ dx } +y-x+xy\quad cotx=0$
$x\frac { dy }{ dx } +(1+xcot x)y=x$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9

Question 10.
$(x+y)\frac { dy }{ dx } =1$
Solution:
$(x+y)\frac { dy }{ dx } =1$
$\frac { 1 }{ (x+y) } \frac { dx }{ dy } =1\Rightarrow \frac { dx }{ dy } =x+y$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10

Question 11.
$ydx+(x-{ y }^{ 2 })dy=0$
Solution:
$ydx+(x-{ y }^{ 2 })dy=0$
$\Rightarrow y\frac { dx }{ dy } +x-{ y }^{ 2 }=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 11

Question 12.
$\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)$
Solution:
$y\frac { dx }{ dy } =x+{ 3y }^{ 2 }\quad or\quad \frac { dx }{ dy } -\frac { x }{ y } =3y$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12

For each of the following Questions 13 to is find a particular solution, satisfying the given condition:

Question 13.
$\frac { dy }{ dx } +2ytanx=sinx,y=0\quad when\quad x=\frac { \pi }{ 3 }$
Solution:
$\frac { dy }{ dx } +(2tanx)y=sinx,P=2tanx$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Question 14.
$\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } +2xy=\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\quad when\quad x=1$
Solution:
$\frac { dy }{ dx } +\frac { 2x }{ 1+{ x }^{ 2 } } y=\frac { 1 }{ { \left( { 1+x }^{ 2 } \right) }^{ 2 } }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 14

Question 15.
$\frac { dy }{ dx } -3ycotx=sin2x,y=2\quad when\quad x=\frac { \pi }{ 2 }$
Solution:
Here P = -3cot x
Q = sin 2x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 15

Question 16.
Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point
Solution:
$\frac { dy }{ dx } =x+y\Rightarrow \frac { dy }{ dx } -y=x\Rightarrow P=-1$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 16

Question 17.
Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5
Solution:
By the given condition
$x+y-\left| \frac { dy }{ dx } \right|=5$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 17

Question 18.
The integrating factor of the differential equation $x\frac { dy }{ dx } -y={ 2x }^{ 2 }$
(a) ${ e }^{ -x }$
(b) ${ e }^{ -y }$
(c) $\frac { 1 }{ x }$
(d) x
Solution:
(c) $P=\frac { -1 }{ x } \therefore IF={ e }^{ -\int { \frac { 1 }{ x } dx } }={ e }^{ -logx }=\frac { 1 }{ x }$

Question 19.
The integrating factor of the differential equation $\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay$ (-1<y<1) is
(a) $\frac { 1 }{ { y }^{ 2 }-1 }$
(b) $\frac { 1 }{ \sqrt { { y }^{ 2 }-1 } }$
(c) $\frac { 1 }{ 1-{ y }^{ 2 } }$
(d) $\frac { 1 }{ \sqrt { { 1-y }^{ 2 } } }$
Solution:
(d) $\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 19

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

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