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 Ex 9.6 Q1.1

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 Ex 9.6 Q2.1
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 Ex 9.6 Q3.1

Ex 9.6 Class 12 Maths Question 4.
$Find the general solution of the following differential equations in Q.1 to 12$
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 = ${ e }^{ \int { { sec }^{ 2 }xdx } }={ e }^{ tanx }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Q5.1

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 Ex 9.6 Q6.1

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 Ex 9.6 Q7.1

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 Ex 9.6 Q8.1

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 Ex 9.6 Q9.1

Ex 9.6 Class 12 Maths Question 10.
$(x+y)\frac { dy }{ dx } =1$
Solution:
$(x+y)\frac { dy }{ dx } =1$
$\frac { 1 }{ (x+y) } \frac { dx }{ dy } =1\Rightarrow \frac { dx }{ dy } =x+y$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Q10.1

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 Ex 9.6 Q11.1

Ex 9.6 Class 12 Maths Question 12.
$Find the general solution of the following differential equations in Q.1 to 12$ 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 Ex 9.6 Q12.1

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 Ex 9.6 Q13.1

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:
$\frac { dy }{ dx } +\frac { 2x }{ 1+{ x }^{ 2 } } y=\frac { 1 }{ { \left( { 1+x }^{ 2 } \right) }^{ 2 } }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Q14.1

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 Ex 9.6 Q15.1

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 Ex 9.6 Q16.1

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 Ex 9.6 Q17.1

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) $\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Q19.1

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