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

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Questions and Answers are prepared by our highly skilled subject experts.

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

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

Question 1.
\(\frac { dy }{ dx }\) + 2y = sin x
Solution:
\(\frac { dy }{ dx }\) + 2y = sin x is a linear differential equation.
∴ p = 2 and Q = sin x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 1
is the general solution.

Question 2.
\(\frac { dy }{ dx } +3y={ e }^{ -2x }\)
Solution:
\(\frac { dy }{ dx } +3y={ e }^{ -2x }\) is a linear differential equation.
∴ p = 3 and Q = e-2x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 2
y = e-2x + Ce-3x is the general solution.

Question 3.
\(\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 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
Required solution is
∴ y(secx + tanx) = (secx + tanx) – x + c

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

Question 5.
\(\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)\)
Solution:
\(\cos ^{2} x \frac{d y}{d x}+y=\tan x\)
i.e., \(\frac{d y}{d x}+y \sec ^{2} x=\sec ^{2} x \tan x\) is a linear differential equation.
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 4
y = (tan x – 1) + Cr-tanx is the general solution.

Question 6.
\(x \frac{d y}{d x}+2 y=x^{2} \log x\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 5
y = \(\frac{x^{2}}{16}(4 \log x-1)+C x^{-2}\) is the general solution.

Question 7.
\(xlogx\frac { dy }{ dx } + y=\frac { 2 }{ x } logx\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 6

Question 8.
(1 + x²)dy + 2xy dx = cotx dx(x ≠ 0)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 7

Question 9.
\(x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 8

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

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

Question 11.
y dx + (x – y²)dy = 0
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 10
is the general solution.

Question 12.
\(\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 11

Question 13.
\(\frac { dy }{ dx } +2ytanx=sinx,y=0\quad when\quad x=\frac { \pi }{ 3 } \)
Solution:
\(\frac { dy }{ dx }\) + 2y tanx = sin x is a linear differential equation.
∴ p = 2 tanx, Q = sin x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 12

Question 14.
\(\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } +2xy=\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\quad when\quad x=1\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 13

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

Question 15.
\(\frac { dy }{ dx } -3ycotx=sin2x,y=2\quad when\quad x=\frac { \pi }{ 2 } \)
Solution:
\(\frac { dy }{ dx }\) – 3y cot x = sin 2x is a linear differential equation.
∴ p = – 3 cot x, Q = sin 2x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 14
y cosec³x = – 2 cosec + C … (1)
When x = \(\frac { π }{ 2 }\), y = 2
(1) → 2 x 1 = – 2 + C ∴ C = 4
Hence the particular solution is
y cosec³x = – 2 cosec + 4
y = \(\frac{-2}{cosec^{2} x}\) + \(\frac{4}{cosec^{3} x}\)
y = 4 sin³x – 2 sin²x

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
\(\frac { dy }{ dx }\) – y = x, is a linear differential equation
∴ P = – 1, Q = x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 15
The curve is passing through the origin
∴ x = 0, y = 0
(1) → C = 1
i.e., x + y + 1 = ex is the required equation of the curve.

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

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:
x + y = \(\frac { dy }{ dx }\)
\(\frac { dy }{ dx }\) – y = x – 5, is a linear differential equation
∴ P = – 1, Q = x – 5
∫P dx = ∫-1 dx = – x
I.F = e∫pdx = e-x
∴ Solution of the differential equation is
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 16
Given that the cure pass through the point (0, 2)
(1) → 2 = 4 + C ∴ C = – 2
y = 4 – x – 2ex is the required equation of the curve.

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:
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 17

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) \(\frac { 1 }{ \sqrt { { 1-y }^{ 2 } } } \)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 18

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