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
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
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
y = e-2x + Ce-3x is the general solution.
Question 3.
\(\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }\)
Solution:
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
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.
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:
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:
Question 8.
(1 + x²)dy + 2xy dx = cotx dx(x ≠ 0)
Solution:
Question 9.
\(x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)\)
Solution:
Question 10.
\((x+y)\frac { dy }{ dx }\) = 1
Solution:
Question 11.
y dx + (x – y²)dy = 0
Solution:
is the general solution.
Question 12.
\(\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)\)
Solution:
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
Question 14.
\(\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } +2xy=\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\quad when\quad x=1\)
Solution:
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
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
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.
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
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:
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) \(\frac { 1 }{ \sqrt { { 1-y }^{ 2 } } } \)