Exercise 9.3 Practice
Chapter 9: Differential Equations – NCERT Solutions (Rationalised)
Topic: Methods of Solving First Order, First Degree Differential Equations (Variable Separable)
Q1
00:00
Find the general solution: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$
$\frac{dy}{dx} = \frac{2\sin^2(x/2)}{2\cos^2(x/2)} = \tan^2(x/2)$.
$dy = (\sec^2(x/2) - 1) dx$. Integrate both sides.
$\int dy = \int (\sec^2(x/2) - 1) dx$.
$\boxed{y = 2\tan(x/2) - x + C}$
Q2
00:00
Find the general solution: $\frac{dy}{dx} = \sqrt{4-y^2}$ ($-2 < y < 2$)
$\frac{dy}{\sqrt{4-y^2}} = dx$. Integrate both sides.
$\sin^{-1}(\frac{y}{2}) = x + C$.
$\boxed{y = 2\sin(x+C)}$
Q3
00:00
Find the general solution: $\frac{dy}{dx} + y = 1$ ($y \ne 1$)
$\frac{dy}{dx} = 1-y \Rightarrow \frac{dy}{1-y} = dx$.
Integrate: $-\log|1-y| = x + C$.
$1-y = e^{-(x+C)} = A e^{-x}$.
$\boxed{y = 1 + A e^{-x}}$
Q4
00:00
Find the general solution: $\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$
Divide by $\tan x \tan y$: $\frac{\sec^2 x}{\tan x} dx + \frac{\sec^2 y}{\tan y} dy = 0$.
Integrate: $\log|\tan x| + \log|\tan y| = \log C$.
$\boxed{\tan x \tan y = C}$
Q5
00:00
Find the general solution: $(e^x + e^{-x}) dy - (e^x - e^{-x}) dx = 0$
$dy = \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$.
Integrate: $y = \log(e^x + e^{-x}) + C$.
$\boxed{y = \log(e^x + e^{-x}) + C}$
Q6
00:00
Find the general solution: $\frac{dy}{dx} = (1+x^2)(1+y^2)$
$\frac{dy}{1+y^2} = (1+x^2) dx$.
Integrate: $\tan^{-1} y = x + \frac{x^3}{3} + C$.
$\boxed{\tan^{-1} y = x + \frac{x^3}{3} + C}$
Q7
00:00
Find the general solution: $y \log y dx - x dy = 0$
$\frac{dx}{x} = \frac{dy}{y \log y}$.
Integrate: $\log|x| = \log|\log y| + \log C$.
$x = C \log y \Rightarrow \log y = \frac{x}{C}$.
$\boxed{y = e^{Cx}}$
Q8
00:00
Find the general solution: $x^5 \frac{dy}{dx} = -y^5$
$y^{-5} dy = -x^{-5} dx$.
Integrate: $\frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C$.
$x^{-4} + y^{-4} = 4C$.
$\boxed{x^{-4} + y^{-4} = k}$
Q9
00:00
Find the general solution: $\frac{dy}{dx} = \sin^{-1} x$
$dy = \sin^{-1} x dx$. Integrate by parts ($1 \cdot \sin^{-1} x$).
$y = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx$.
$\boxed{y = x \sin^{-1} x + \sqrt{1-x^2} + C}$
Q10
00:00
Find the general solution: $e^x \tan y dx + (1-e^x) \sec^2 y dy = 0$
$\frac{e^x}{1-e^x} dx + \frac{\sec^2 y}{\tan y} dy = 0$.
Integrate: $-\log|1-e^x| + \log|\tan y| = \log C$.
$\boxed{\tan y = C(1-e^x)}$
Q11
00:00
Find particular solution: $(x^3+x^2+x+1)\frac{dy}{dx} = 2x^2+x; y=1$ when $x=0$
$dy = \frac{2x^2+x}{(x+1)(x^2+1)} dx$. Use partial fractions.
$\frac{2x^2+x}{(x+1)(x^2+1)} = \frac{1}{2(x+1)} + \frac{3x-1}{2(x^2+1)}$.
Integrate and apply $y(0)=1$ to find $C$.
$\boxed{y = \frac{1}{4}\log(x+1)^2(x^2+1)^3 - \frac{1}{2}\tan^{-1}x + 1}$
Q12
00:00
Find particular solution: $x(x^2-1)\frac{dy}{dx} = 1; y=0$ when $x=2$
$dy = \frac{dx}{x(x-1)(x+1)}$. Partial fractions.
$y = -\log x + \frac{1}{2}\log(x-1) + \frac{1}{2}\log(x+1) + C$.
Apply $y(2)=0 \Rightarrow C = \log(2/\sqrt{3})$.
$\boxed{y = \frac{1}{2}\log|\frac{x^2-1}{x^2}| - \frac{1}{2}\log \frac{3}{4}}$
Q13
00:00
Find particular solution: $\cos(\frac{dy}{dx}) = a; y=1$ when $x=0$
$\frac{dy}{dx} = \cos^{-1} a \Rightarrow dy = \cos^{-1} a \cdot dx$.
$y = x \cos^{-1} a + C$. Apply $y(0)=1 \Rightarrow C=1$.
$\boxed{y = x \cos^{-1} a + 1 \text{ or } \cos(\frac{y-1}{x}) = a}$
Q14
00:00
Find particular solution: $\frac{dy}{dx} = y \tan x; y=1$ when $x=0$
$\frac{dy}{y} = \tan x dx$.
$\log|y| = \log|\sec x| + C$.
$y(0)=1 \Rightarrow 0 = 0 + C \Rightarrow C=0$.
$\boxed{y = \sec x}$
Q15
00:00
Find equation of curve passing through $(0,0)$ and $y' = e^x \sin x$.
$y = \int e^x \sin x dx$. Use integration by parts twice.
$y = \frac{e^x}{2}(\sin x - \cos x) + C$.
Passes through $(0,0) \Rightarrow 0 = -1/2 + C \Rightarrow C = 1/2$.
$\boxed{y = \frac{e^x}{2}(\sin x - \cos x) + \frac{1}{2}}$
Q16
00:00
For the differential equation $xy \frac{dy}{dx} = (x+2)(y+2)$, find the solution curve passing through $(1, -1)$.
$\frac{y}{y+2} dy = \frac{x+2}{x} dx \Rightarrow (1 - \frac{2}{y+2}) dy = (1 + \frac{2}{x}) dx$.
$y - 2\log|y+2| = x + 2\log|x| + C$.
At $(1, -1): -1 - 0 = 1 + 0 + C \Rightarrow C = -2$.
$\boxed{y - x + 2 = \log(x^2(y+2)^2)}$
Q17
00:00
Find the equation of a curve passing through $(0, -2)$ given that at any point $(x, y)$, the product of the slope of its tangent and y-coordinate is equal to the x-coordinate.
$y \cdot \frac{dy}{dx} = x \Rightarrow y dy = x dx$.
$\frac{y^2}{2} = \frac{x^2}{2} + C$.
At $(0, -2): 2 = 0 + C \Rightarrow C = 2$.
$\boxed{y^2 - x^2 = 4}$
Q18
00:00
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 $(-4, -3)$. Find the equation of the curve given that it passes through $(-2, 1)$.
$\frac{dy}{dx} = 2 \frac{y+3}{x+4}$.
$\frac{dy}{y+3} = \frac{2dx}{x+4} \Rightarrow \log|y+3| = 2\log|x+4| + \log C$.
$y+3 = C(x+4)^2$. At $(-2, 1): 4 = C(2)^2 \Rightarrow C=1$.
$\boxed{y+3 = (x+4)^2}$
Q19
00:00
The volume of a 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.
$\frac{dV}{dt} = k \Rightarrow \frac{d}{dt}(\frac{4}{3}\pi r^3) = k \Rightarrow 4\pi r^2 \frac{dr}{dt} = k$.
Integrate: $\frac{4}{3}\pi r^3 = kt + C$.
At $t=0, r=3 \Rightarrow C = 36\pi$. At $t=3, r=6 \Rightarrow k = 84\pi$.
$\boxed{r = (63t + 27)^{1/3}}$
Q20
00:00
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 ($\log_e 2 = 0.6931$).
$\frac{dP}{dt} = \frac{rP}{100} \Rightarrow \frac{dP}{P} = \frac{r}{100} dt$.
$\log P = \frac{rt}{100} + C$. At $t=0, P=100 \Rightarrow C = \log 100$.
At $t=10, P=200 \Rightarrow \log 2 = \frac{10r}{100}$.
$\boxed{r = 6.93\%}$
Q21
00:00
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 ($e^{0.5} = 1.648$).
$\frac{dP}{dt} = 0.05P \Rightarrow P = C e^{0.05t}$.
At $t=0, P=1000 \Rightarrow C=1000$.
At $t=10, P = 1000 e^{0.5} = 1000(1.648)$.
$\boxed{\text{Rs } 1648}$
Q22
00:00
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?
$\frac{dy}{dt} = ky \Rightarrow y = C e^{kt}$.
$t=0, y=100000 \Rightarrow C=100000$.
$t=2, y=110000 \Rightarrow e^{2k} = 1.1 \Rightarrow k = \frac{1}{2}\log(1.1)$.
For $y=200000 \Rightarrow 2 = e^{kt} \Rightarrow t = \frac{\log 2}{k}$.
$\boxed{\frac{2\log 2}{\log 1.1} \text{ hours}}$
Q23
00:00
The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is:
(A) $e^x + e^{-y} = C$ (B) $e^x + e^y = C$ (C) $e^{-x} + e^y = C$ (D) $e^{-x} + e^{-y} = C$
(A) $e^x + e^{-y} = C$ (B) $e^x + e^y = C$ (C) $e^{-x} + e^y = C$ (D) $e^{-x} + e^{-y} = C$
$\frac{dy}{dx} = e^x e^y \Rightarrow e^{-y} dy = e^x dx$.
$-e^{-y} = e^x + k \Rightarrow e^x + e^{-y} = C$.
$\boxed{\text{(A) } e^x + e^{-y} = C}$