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.
- Differential Equations Class 12 Ex 9.1
- Differential Equations Class 12 Ex 9.2
- Differential Equations Class 12 Ex 9.3
- Differential Equations Class 12 Ex 9.4
- Differential Equations Class 12 Ex 9.5
| 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
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.
\(\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)\)
Solution:
\(y\frac { dx }{ dy } =x+{ 3y }^{ 2 }\quad or\quad \frac { dx }{ dy } -\frac { x }{ y } =3y\)
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) \(\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay\)
We hope the NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 help you. If you have any query regarding NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6, drop a comment below and we will get back to you at the earliest.