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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 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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 }
byjus class 12 maths Chapter 9 Differential Equations 5

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9.1

Ex 9.6 Class 12 Maths 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
byjus class 12 maths Chapter 9 Differential Equations 10

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12.1

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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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 } }
byjus class 12 maths Chapter 9 Differential Equations 14

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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

Ex 9.6 Class 12 Maths 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
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 17.1

Ex 9.6 Class 12 Maths 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 }

Ex 9.6 Class 12 Maths 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
byjus class 12 maths Chapter 9 Differential Equations 19

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