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

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 pdf, free NCERT solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 book pdf download.

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

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

Ex 9.6 Class 12 Maths Question 2.
$\frac { dy }{ dx } +3y={ e }^{ -2x }$
Solution:
$\frac { dy }{ dx } +3y={ e }^{ -2x }$
Here P = 3, $IF={ e }^{ \int { p.dx } }={ e }^{ 3x }$

which is required equation

Ex 9.6 Class 12 Maths Question 3.
$\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }$
Solution:
$\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }$
$IF={ e }^{ \int { \frac { 1 }{ x } dx } }={ e }^{ logx }=x$

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 = ${ e }^{ \int { { sec }^{ 2 }xdx } }={ e }^{ tanx }$

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

Ex 9.6 Class 12 Maths Question 7.
$xlogx\frac { dy }{ dx } +y=\frac { 2 }{ x } logx$
Solution:
$\frac { dy }{ dx } +\frac { 1 }{ xlogx } y=\frac { 2 }{ { x }^{ 2 } }$

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

Ex 9.6 Class 12 Maths Question 9.
$x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)$
Solution:
$x\frac { dy }{ dx } +y-x+xy\quad cotx=0$
$x\frac { dy }{ dx } +(1+xcot x)y=x$

Ex 9.6 Class 12 Maths Question 10.
$(x+y)\frac { dy }{ dx } =1$
Solution:
$(x+y)\frac { dy }{ dx } =1$
$\frac { 1 }{ (x+y) } \frac { dx }{ dy } =1\Rightarrow \frac { dx }{ dy } =x+y$

Ex 9.6 Class 12 Maths Question 11.
$ydx+(x-{ y }^{ 2 })dy=0$
Solution:
$ydx+(x-{ y }^{ 2 })dy=0$
$\Rightarrow y\frac { dx }{ dy } +x-{ y }^{ 2 }=0$

Ex 9.6 Class 12 Maths Question 12.
$\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)”><br><strong>Solution:</strong><br><img src=$

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:
$\frac { dy }{ dx } +(2tanx)y=sinx,P=2tanx$

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:
$\frac { dy }{ dx } +\frac { 2x }{ 1+{ x }^{ 2 } } y=\frac { 1 }{ { \left( { 1+x }^{ 2 } \right) }^{ 2 } }$

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

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:
$\frac { dy }{ dx } =x+y\Rightarrow \frac { dy }{ dx } -y=x\Rightarrow P=-1$

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
$x+y-\left| \frac { dy }{ dx } \right|=5$

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) $\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay$

All Chapter NCERT Solutions For Class12 Maths

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All Subject NCERT Solutions For Class12

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

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