NCERT Solutions Class-12-Chapter-9-Differential Equations
Excercise-9.3
Note:
Variable Separable Method: If a differential equation can be expressed in the form \( f(x)dx = g(y)dy \), it is called a variable separable type.
Solution Steps:
1. Separate the terms of \( x \) and \( y \).
2. Integrate both sides: \( \int f(x)dx = \int g(y)dy + C \).
3. For particular solutions, use the given boundary conditions (e.g., \( y=y_0 \) when \( x=x_0 \)) to find the value of the constant \( C \).
Q1
Find the general solution: \( \frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x} \)
Solution:
Using trigonometric identities: \( 1 - \cos x = 2\sin^2 \frac{x}{2} \) and \( 1 + \cos x = 2\cos^2 \frac{x}{2} \).
\[ \frac{dy}{dx} = \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2} = \sec^2 \frac{x}{2} - 1 \]
Separating variables and integrating:
\[ \int dy = \int \left(\sec^2 \frac{x}{2} - 1\right) dx \]
\[ y = \frac{\tan \frac{x}{2}}{1/2} - x + C \Rightarrow y = 2\tan \frac{x}{2} - x + C \]
Q2
Find the general solution: \( \frac{dy}{dx} = \sqrt{4 - y^2} \quad (-2 < y < 2) \)
Solution:
Separating variables:
\[ \frac{dy}{\sqrt{4 - y^2}} = dx \]
Integrating both sides:
\[ \int \frac{dy}{\sqrt{2^2 - y^2}} = \int dx \]
Using \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1} \frac{x}{a} \):
\[ \sin^{-1} \left(\frac{y}{2}\right) = x + C \]
\[ \frac{y}{2} = \sin(x + C) \Rightarrow y = 2\sin(x + C) \]
Q3
Find the general solution: \( \frac{dy}{dx} + y = 1 \quad (y \neq 1) \)
Solution:
\( \frac{dy}{dx} = 1 - y \Rightarrow \frac{dy}{1-y} = dx \).
Integrating both sides:
\[ \int \frac{dy}{1-y} = \int dx \Rightarrow -\log|1-y| = x + C_1 \]
\( \log|1-y| = -x - C_1 \).
\( 1-y = e^{-x-C_1} = e^{-C_1}e^{-x} = A e^{-x} \) (where \( A = e^{-C_1} \)).
\( y = 1 - A e^{-x} \) or \( y = 1 + A e^{-x} \) (depending on constant definition).
Q4
Find the general solution: \( \sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0 \)
Solution:
\( \sec^2 y \tan x \, dy = -\sec^2 x \tan y \, dx \).
Dividing both sides by \( \tan x \tan y \):
\[ \frac{\sec^2 y}{\tan y} dy = -\frac{\sec^2 x}{\tan x} dx \]
Integrating both sides (put \( \tan y = u, \tan x = v \)):
\( \log|\tan y| = -\log|\tan x| + \log C \).
\( \log|\tan y| + \log|\tan x| = \log C \).
\( \log|\tan x \tan y| = \log C \Rightarrow \tan x \tan y = C \).
Q5
Find the general solution: \( (e^x + e^{-x}) dy - (e^x - e^{-x}) dx = 0 \)
Solution:
\( (e^x + e^{-x}) dy = (e^x - e^{-x}) dx \).
\[ dy = \frac{e^x - e^{-x}}{e^x + e^{-x}} dx \]
Integrating both sides (let \( e^x + e^{-x} = t \Rightarrow (e^x - e^{-x})dx = dt \)):
\[ y = \int \frac{dt}{t} = \log|t| + C \]
\( y = \log(e^x + e^{-x}) + C \).
Q6
Find the general solution: \( \frac{dy}{dx} = (1+x^2)(1+y^2) \)
Solution:
Separating variables:
\[ \frac{dy}{1+y^2} = (1+x^2) dx \]
Integrating:
\[ \tan^{-1} y = x + \frac{x^3}{3} + C \]
Q7
Find the general solution: \( y \log y \, dx - x \, dy = 0 \)
Solution:
\( y \log y \, dx = x \, dy \Rightarrow \frac{dx}{x} = \frac{dy}{y \log y} \).
Integrating both sides (Put \( \log y = t \Rightarrow \frac{1}{y}dy = dt \)):
\[ \int \frac{dx}{x} = \int \frac{dt}{t} \Rightarrow \log|x| = \log|\log y| + \log C \]
\( \log x = \log(C \log y) \Rightarrow x = C \log y \Rightarrow \log y = \frac{x}{C} \).
\( y = e^{x/C} \) or \( y = e^{Ax} \).
Q8
Find the general solution: \( x^5 \frac{dy}{dx} = -y^5 \)
Solution:
Separating variables:
\[ \frac{dy}{y^5} = -\frac{dx}{x^5} \Rightarrow y^{-5} dy = -x^{-5} dx \]
Integrating:
\[ \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C \Rightarrow -\frac{1}{4y^4} = \frac{1}{4x^4} + C \]
\( x^{-4} + y^{-4} = k \) (where \( k = -4C \)).
Q9
Find the general solution: \( \frac{dy}{dx} = \sin^{-1} x \)
Solution:
\( dy = \sin^{-1} x \, dx \).
Integrate using integration by parts (\( u = \sin^{-1}x, v = 1 \)):
\[ y = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx \]
Put \( 1-x^2 = t^2 \Rightarrow -2x dx = 2t dt \Rightarrow x dx = -t dt \).
\[ y = x \sin^{-1} x - \int \frac{-t dt}{t} = x \sin^{-1} x + \int dt \]
\( y = x \sin^{-1} x + t + C = x \sin^{-1} x + \sqrt{1-x^2} + C \).
Q10
Find the general solution: \( e^x \tan y \, dx + (1-e^x) \sec^2 y \, dy = 0 \)
Solution:
Separating variables:
\[ \frac{e^x}{1-e^x} dx = -\frac{\sec^2 y}{\tan y} dy \]
Integrating (Put \( 1-e^x = u \) and \( \tan y = v \)):
\( -\int \frac{du}{u} = -\int \frac{dv}{v} \Rightarrow \log|u| = \log|v| + \log C \).
\( \log|1-e^x| = \log|C \tan y| \).
\( 1-e^x = C \tan y \) or \( \tan y = k(1-e^x) \).
Q11
Find particular solution: \( (x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x \); \( y=1 \) when \( x=0 \).
Solution:
Factorize denominator: \( x^3+x^2+x+1 = x^2(x+1) + 1(x+1) = (x+1)(x^2+1) \).
\( dy = \frac{2x^2+x}{(x+1)(x^2+1)} dx \).
Partial fractions: \( \frac{2x^2+x}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1} \). Solving yields \( A=1/2, B=3/2, C=-1/2 \).
\( y = \int \left( \frac{1/2}{x+1} + \frac{\frac{3}{2}x - \frac{1}{2}}{x^2+1} \right) dx \).
\( y = \frac{1}{2}\log|x+1| + \frac{3}{4}\log(x^2+1) - \frac{1}{2}\tan^{-1}x + C \).
Using \( y(0)=1 \): \( 1 = 0 + 0 - 0 + C \Rightarrow C=1 \).
Solution: \( y = \frac{1}{4}\log(x+1)^2(x^2+1)^3 - \frac{1}{2}\tan^{-1}x + 1 \).
Q12
Find particular solution: \( x(x^2-1)\frac{dy}{dx} = 1 \); \( y=0 \) when \( x=2 \).
Solution:
\( dy = \frac{dx}{x(x-1)(x+1)} \).
Partial fractions: \( \frac{1}{x(x^2-1)} = -\frac{1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1} \).
\( y = -\log|x| + \frac{1}{2}\log|x-1| + \frac{1}{2}\log|x+1| + C \).
\( y = \frac{1}{2} \log \left| \frac{x^2-1}{x^2} \right| + C \).
At \( x=2, y=0 \): \( 0 = \frac{1}{2} \log \frac{3}{4} + C \Rightarrow C = -\frac{1}{2} \log \frac{3}{4} \).
Solution: \( y = \frac{1}{2} \log \left| \frac{x^2-1}{x^2} \right| - \frac{1}{2} \log \frac{3}{4} \).
Q13
Find particular solution: \( \cos\left(\frac{dy}{dx}\right) = a \quad (a \in \mathbb{R}) \); \( y=1 \) when \( x=0 \).
Solution:
\( \frac{dy}{dx} = \cos^{-1} a \Rightarrow dy = (\cos^{-1} a) dx \).
\( y = x \cos^{-1} a + C \).
At \( x=0, y=1 \): \( 1 = 0 + C \Rightarrow C=1 \).
\( y = x \cos^{-1} a + 1 \Rightarrow \frac{y-1}{x} = \cos^{-1} a \Rightarrow \cos\left(\frac{y-1}{x}\right) = a \).
Q14
Find particular solution: \( \frac{dy}{dx} = y \tan x \); \( y=1 \) when \( x=0 \).
Solution:
\( \frac{dy}{y} = \tan x \, dx \).
\( \log|y| = \log|\sec x| + C \).
At \( x=0, y=1 \): \( \log 1 = \log 1 + C \Rightarrow 0 = 0 + C \Rightarrow C = 0 \).
\( \log|y| = \log|\sec x| \Rightarrow y = \sec x \).
Q15
Find the equation of a curve passing through the point (0, 0) and whose differential equation is \( y' = e^x \sin x \).
Solution:
\( dy = e^x \sin x \, dx \).
Using integration by parts for \( I = \int e^x \sin x \, dx \):
\( I = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - (e^x \cos x - \int e^x (-\sin x) dx) \).
\( I = e^x(\sin x - \cos x) - I \Rightarrow 2I = e^x(\sin x - \cos x) \).
\( y = \frac{e^x}{2}(\sin x - \cos x) + C \).
Passes through (0,0): \( 0 = \frac{1}{2}(0-1) + C \Rightarrow C = 1/2 \).
Equation: \( y = \frac{e^x}{2}(\sin x - \cos x) + \frac{1}{2} \) or \( 2y = e^x(\sin x - \cos x) + 1 \).
Q16
Find the solution curve for \( xy \frac{dy}{dx} = (x+2)(y+2) \) passing through (1, -1).
Solution:
Separating variables: \( \frac{y}{y+2} dy = \frac{x+2}{x} dx \).
\( \int \left( 1 - \frac{2}{y+2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx \).
\( y - 2\log|y+2| = x + 2\log|x| + C \).
Using (1, -1): \( -1 - 2\log 1 = 1 + 2\log 1 + C \Rightarrow -1 = 1 + C \Rightarrow C = -2 \).
Equation: \( y - x + 2 = 2\log|x(y+2)| \).
Q17
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate is equal to the x-coordinate.
Solution:
Condition: \( y \cdot \frac{dy}{dx} = x \).
\( y \, dy = x \, dx \).
Integrating: \( \frac{y^2}{2} = \frac{x^2}{2} + C_1 \Rightarrow y^2 - x^2 = 2C_1 = k \).
Passes through (0, -2): \( (-2)^2 - 0 = k \Rightarrow k = 4 \).
Equation: \( y^2 - x^2 = 4 \).
Q18
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve given that it passes through (-2, 1).
Solution:
Slope of line joining \( (x,y) \) and \( (-4,-3) \) is \( \frac{y+3}{x+4} \).
Differential equation: \( \frac{dy}{dx} = 2 \left( \frac{y+3}{x+4} \right) \).
\( \frac{dy}{y+3} = \frac{2}{x+4} dx \).
\( \log|y+3| = 2\log|x+4| + \log C = \log(C(x+4)^2) \).
\( y+3 = C(x+4)^2 \).
Passes through (-2, 1): \( 1+3 = C(-2+4)^2 \Rightarrow 4 = 4C \Rightarrow C=1 \).
Equation: \( y+3 = (x+4)^2 \).
Q19
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after \( t \) seconds.
Solution:
Volume \( V = \frac{4}{3}\pi r^3 \). Rate of change \( \frac{dV}{dt} = k \).
\( \frac{d}{dt}(\frac{4}{3}\pi r^3) = k \Rightarrow 4\pi r^2 \frac{dr}{dt} = k \).
\( 4\pi r^2 dr = k dt \Rightarrow \frac{4}{3}\pi r^3 = kt + C \).
At \( t=0, r=3 \): \( \frac{4}{3}\pi (27) = C \Rightarrow C = 36\pi \).
At \( t=3, r=6 \): \( \frac{4}{3}\pi (216) = 3k + 36\pi \Rightarrow 288\pi = 3k + 36\pi \Rightarrow 3k = 252\pi \Rightarrow k = 84\pi \).
Equation: \( \frac{4}{3}\pi r^3 = 84\pi t + 36\pi \).
Simplifying: \( r^3 = 63t + 27 \Rightarrow r = (63t + 27)^{1/3} \).
Q20
In a bank, principal increases continuously at the rate of \( r\% \) per year. Find the value of \( r \) if Rs 100 double itself in 10 years (\( \log_e 2 = 0.6931 \)).
Solution:
\( \frac{dP}{dt} = \frac{r}{100} P \Rightarrow \frac{dP}{P} = \frac{r}{100} dt \).
\( \log P = \frac{rt}{100} + C \).
At \( t=0, P=P_0 \Rightarrow C = \log P_0 \). So \( \log \frac{P}{P_0} = \frac{rt}{100} \).
Here \( P_0 = 100, P = 200, t = 10 \).
\( \log 2 = \frac{r(10)}{100} = \frac{r}{10} \).
\( r = 10 \log 2 = 10(0.6931) = 6.931 \).
Q21
In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (\( e^{0.5} = 1.648 \)).
Solution:
From previous formula: \( P = P_0 e^{rt/100} \).
\( P_0 = 1000, r = 5, t = 10 \).
\( P = 1000 e^{(5 \times 10)/100} = 1000 e^{0.5} \).
\( P = 1000 \times 1.648 = 1648 \).
Amount = Rs 1648.
Q22
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Solution:
\( \frac{dy}{dt} = ky \Rightarrow y = C e^{kt} \).
At \( t=0, y=100000 \Rightarrow C = 100000 \). So \( y = 100000 e^{kt} \).
At \( t=2, y=110000 \) (10% increase): \( 110000 = 100000 e^{2k} \Rightarrow e^{2k} = 1.1 \).
\( 2k = \log 1.1 \Rightarrow k = \frac{1}{2} \log 1.1 \).
Find \( t \) when \( y = 200000 \):
\( 200000 = 100000 e^{kt} \Rightarrow 2 = e^{kt} \Rightarrow kt = \log 2 \).
\( t = \frac{\log 2}{k} = \frac{\log 2}{\frac{1}{2} \log 1.1} = \frac{2 \log 2}{\log 1.1} \) hours.
Q23
The general solution of the differential equation \( \frac{dy}{dx} = e^{x+y} \) is- \( e^x + e^{-y} = C \)
- \( e^x + e^y = C \)
- \( e^{-x} + e^y = C \)
- \( e^{-x} + e^{-y} = C \)
- \( e^x + e^{-y} = C \)
- \( e^x + e^y = C \)
- \( e^{-x} + e^y = C \)
- \( e^{-x} + e^{-y} = C \)
Answer: (A)
Solution:
\( \frac{dy}{dx} = e^x \cdot e^y \Rightarrow e^{-y} dy = e^x dx \).
Integrating: \( -e^{-y} = e^x + K \).
\( e^x + e^{-y} = -K = C \).