Anand Classes offers a dedicated PDF for NCERT Solutions of Class 12 Maths Chapter 9 – Differential Equations Exercise 9.3 (Set-2), specially designed to assist students in mastering the key concepts of first-order, first-degree differential equations, integration techniques and application-based problems. The detailed step-by-step solutions follow the latest CBSE and NCERT syllabus format, making them ideal for rigorous exam preparation and revision. Click the print button to download study material and notes.
NCERT Question.13 : Find a particular solution satisfying the given condition:
$$\cos\left(\frac{dy}{dx}\right)=a\qquad(a\in\mathbb{R}),\qquad y(0)=1$$
Solution :
From the equation
$$\cos\left(\frac{dy}{dx}\right)=a$$
we get (principal value)
$$\frac{dy}{dx}=\cos^{-1}a.$$
Note: for a real principal value $\cos^{-1}a$ we require $a\in[-1,1]$; if $a$ lies outside $[-1,1]$ the derivative is not real.
Integrate with respect to $x$:
$$y=\int\cos^{-1}adx = x\cos^{-1}a + C.$$
Use the initial condition $y(0)=1$:
$$1=0\cdot\cos^{-1}a + C \quad\Longrightarrow\quad C=1.$$
Final Answer
$$\boxed{y=x\cos^{-1}a+1\qquad (a\in[-1,1],\text{for a real solution})}$$
Grab detailed notes from Anand Classes — perfect for JEE & CBSE revision, concise and high-quality practice material to strengthen your first-order differential equations skills.
NCERT Question.14 : Find a particular solution satisfying the given condition:
$$\frac{dy}{dx}=y\tan x;\qquad y=1\ \text{when}\ x=0$$
Solution:
Rewrite the equation as
$$\frac{dy}{y}=\tan xdx.$$
Integrate both sides:
$$\int\frac{dy}{y}=\int\tan xdx.$$
Hence
$$\ln|y|=\ln|\sec x|+C.$$
Exponentiating and absorbing the constant gives
$$y=C\sec x.$$
Use the condition $y(0)=1$:
$$1=C\sec 0\implies C=1.$$
Final Answer
$$\boxed{y=\sec x}$$
Get concise revision notes by Anand Classes — perfect for JEE and CBSE practice, a clear resource for mastering first-order differential equations.
NCERT Question.15 : Find the equation of a curve passing through $(0, 0)$ and whose differential equation is $y’ = e^{x} \sin x$
Solution :
The given differential equation is
$$
\frac{dy}{dx} = e^{x} \sin x
$$
Separating variables and integrating both sides:
$$
y = \int e^{x} \sin x dx
$$
Let
$$ I = \int e^{x} \sin x dx $$
Using integration by parts:
$$
I = e^{x} \sin x – \int e^{x} \cos x dx
$$
Let
$$ J = \int e^{x} \cos x dx $$
Then
$$
J = e^{x} \cos x + \int e^{x} \sin x dx = e^{x} \cos x + I
$$
Substitute the value of $J$ into the previous equation:
$$
I = e^{x} \sin x – (e^{x} \cos x + I)
$$
$$
2I = e^{x} (\sin x – \cos x)
$$
$$
I = \frac{e^{x}}{2} (\sin x – \cos x)
$$
Hence, the general solution is
$$
y = \frac{e^{x}}{2} (\sin x – \cos x) + C
$$
Now, using the condition $y = 0$ when $x = 0$:
$$
0 = \frac{e^{0}}{2} (\sin 0 – \cos 0) + C
$$
$$
0 = \frac{1}{2} (0 – 1) + C
$$
$$
C = \frac{1}{2}
$$
Therefore, the required equation is
$$
\boxed{y = \frac{e^{x}}{2} (\sin x – \cos x) + \frac{1}{2}}
$$
or equivalently,
$$
\boxed{2y – 1 = e^{x} (\sin x – \cos x)}
$$
Download detailed JEE and CBSE notes by Anand Classes — the most concise material for mastering first-order differential equations and integration by parts.
NCERT Question.16 : For the differential equation
$$xy\frac{dy}{dx}=(x+2)(y+2)$$
Find the solution curve passing through the point $(1, –1)$.
Solution :
Rewrite as
$$\frac{y}{y+2}dy=\frac{x+2}{x}dx.$$
Simplify the integrands:
$$\frac{y}{y+2}=1-\frac{2}{y+2},\qquad \frac{x+2}{x}=1+\frac{2}{x}.$$
Integrate both sides:
$$\int\Big(1-\frac{2}{y+2}\Big)dy=\int\Big(1+\frac{2}{x}\Big)dx.$$
This gives
$$y-2\ln|y+2|=x+2\ln|x|+C.$$
Use the point $(1,-1)$ to find $C$:
$$-1-2\ln|1|=1+2\ln|1|+C\quad\Longrightarrow\quad -1=1+C\quad\Longrightarrow\quad C=-2.$$
Hence the particular solution (implicit) is
$$y-2\ln|y+2|=x+2\ln|x|-2.$$
Equivalently,
$$y-x+2=2\ln\big|x(y+2)\big|\quad\text{or}\quad \ln\big(x^2(y+2)^2\big)=y-x+2.$$
Final Answer : Thus
$$\boxed{\ln\big(x^2(y+2)^2\big)=y-x+2}$$
is the required equation
Download concise notes by Anand Classes — perfect for JEE and CBSE revision and quick practice on separable differential equations.
NCERT Question.17 : Find the equation of the curve passing through $(0,-2)$ given that at any point $(x,y)$ the product of the slope of the tangent and the $y$–coordinate equals the $x$–coordinate.
Solution :
The condition is
$$y\frac{dy}{dx}=x.$$
Separate variables:
$$ydy=xdx.$$
Integrate both sides:
$$\int ydy=\int xdx\quad\Longrightarrow\quad \frac{y^{2}}{2}=\frac{x^{2}}{2}+C.$$
Multiply by $2$ and rearrange:
$$y^{2}-x^{2}=2C.$$
Use the point $(0,-2)$:
$$(-2)^{2}-0^{2}=2C\implies 4=2C\implies C=2.$$
Final Answer : Thus,
$$\boxed{y^{2}-x^{2}=4,}$$
is the required equation.
Download practice notes by Anand Classes — perfect for JEE and CBSE revision and a clear summary of separable differential equations.
NCERT Question.18 : At any point $(x,y)$ of a curve, the slope of the tangent is twice the slope of the line joining the point to $(-4,-3)$. Find the equation of the curve given that it is passes through $(-2,1)$.
Solution :
Slope of the line joining $(x,y)$ to $(-4,-3)$ is
$$\frac{y-(-3)}{x-(-4)}=\frac{y+3}{x+4}.$$
Since the tangent slope is twice this,
$$\frac{dy}{dx}=2\cdot\frac{y+3}{x+4}.$$
Separate variables:
$$\frac{dy}{y+3}=2\frac{dx}{x+4}.$$
Integrate both sides:
$$\int\frac{dy}{y+3}=2\int\frac{dx}{x+4}$$
$$\ln|y+3|=2\ln|x+4|+C.$$
Exponentiate:
$$y+3=C'(x+4)^2$$
where $(C’=e^{C})$. Use the point $(-2,1)$:
$$1+3=C'( -2+4)^2\quad\Longrightarrow\quad 4=C’\cdot 2^2\quad\Longrightarrow\quad C’=1.$$
Final Answer
$$\boxed{y+3=(x+4)^2}$$
Download practice notes by Anand Classes — perfect for JEE and CBSE revision and concise problem-solving practice.
NCERT Question.19 : The volume of a spherical balloon 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 :
The volume of a sphere is
$$
V = \frac{4}{3} \pi r^3
$$
Given that
$$
\frac{dV}{dt} = k
$$
where $k$ is constant. Integrating both sides,
$$
V = kt + C
$$
When $t = 0$, $r = 3$:
$$
V = \frac{4}{3} \pi (3)^3 = 36\pi
$$
Hence, $C = 36\pi$.
When $t = 3$, $r = 6$:
$$
V = \frac{4}{3} \pi (6)^3 = 288\pi
$$
So,
$$
288\pi = 3k + 36\pi
$$
$$
3k = 252\pi
$$
$$
k = 84\pi
$$
Thus,
$$
V = 84\pi t + 36\pi
$$
Now,
$$
\frac{4}{3}\pi r^3 = 84\pi t + 36\pi
$$
$$
r^3 = \frac{3}{4\pi}(84\pi t + 36\pi)
$$
$$
r^3 = 63t + 27
$$
Final Answer
$$
\boxed{r = (63t + 27)^{\dfrac{1}{3}}}
$$
Download comprehensive math notes and solved differential equations by Anand Classes — ideal for JEE, CBSE, and advanced calculus preparation.
NCERT Question.20 : In a bank, the principal increases continuously at the rate of $r%$ per year. Find the value of $r$ if Rs 100 doubles itself in 10 years (given $\log_e 2 = 0.6931$).
Solution :
Given:
$$
\frac{dp}{dt} = \frac{r}{100}p
$$
Rearranging,
$$
\frac{dp}{p} = \frac{r}{100}dt
$$
Integrating both sides,
$$
\int \frac{dp}{p} = \frac{r}{100} \int dt
$$
$$
\log p = \frac{rt}{100} + k
$$
$$
p = e^{\dfrac{rt}{100} + k} = e^k \cdot e^{\dfrac{rt}{100}}
$$
Let $e^k = A$, then
$$
p = A e^{\dfrac{rt}{100}}
$$
Now, when $t = 0$, $p = 100$:
$$
100 = A e^0 \Rightarrow A = 100
$$
Thus,
$$
p = 100 e^{\dfrac{rt}{100}}
$$
When $t = 10$, $p = 200$:
$$
200 = 100 e^{\dfrac{r \times 10}{100}}
$$
$$
2 = e^{\dfrac{r}{10}}
$$
Taking log on both sides,
$$
\log_e 2 = \dfrac{r}{10}
$$
$$
r = 10 \times \log_e 2
$$
$$
r = 10 \times 0.6931
$$
$$
\boxed{r = 6.931\%}
$$
Download complete exponential growth and differential equation notes by Anand Classes — perfect for JEE, CBSE, and competitive exam preparation.
NCERT Question.21 : In a bank, the principal increases continuously at the rate of $5%$ per year. An amount of Rs 1000 is deposited in the bank. Find its value after 10 years (given $e^{0.5} = 1.648$).
Solution :
Given:
$$
\frac{dp}{dt} = \frac{5}{100}p
$$
Rearranging,
$$
\frac{dp}{p} = \frac{1}{20}dt
$$
Integrating both sides,
$$
\int \frac{dp}{p} = \frac{1}{20} \int dt
$$
$$
\log p = \frac{t}{20} + C
$$
$$
p = e^{\dfrac{t}{20} + C} = e^C \cdot e^{\dfrac{t}{20}}
$$
Let $e^C = k$, then
$$
p = k e^{\dfrac{t}{20}}
$$
Now, when $t = 0$, $p = 1000$:
$$
1000 = k e^0 \Rightarrow k = 1000
$$
Thus,
$$
p = 1000 e^{\dfrac{t}{20}}
$$
When $t = 10$:
$$
p = 1000 e^{\dfrac{10}{20}} = 1000 e^{0.5}
$$
Substituting $e^{0.5} = 1.648$:
$$
p = 1000 \times 1.648 = 1648
$$
Hence,
$$
\boxed{p = 1648}
$$
This shows that after 10 years, the amount will be Rs 1648.
Get detailed exponential growth and compound interest notes by Anand Classes — ideal for JEE, CBSE, and board exam preparation.
NCERT Question.22 : In a culture the bacteria count is $100000$. 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 :
Let $y(t)$ be the bacteria count and $k$ the constant of proportionality. Then
$$\frac{dy}{dt}=ky\quad\Longrightarrow\quad \frac{dy}{y}=kdt.$$
Integrating,
$$\ln y = kt + C\quad\Longrightarrow\quad y=Ae^{kt}\ \text{(where }A=e^{C}\text{).}$$
From the given: after $2$ hours the population multiplies by $1.1$, so
$$1.1 = e^{2k}\quad\Longrightarrow\quad k=\frac{1}{2}\ln(1.1).$$
To double the population ($y$ becomes $2A$),
$$2 = e^{kt}\quad\Longrightarrow\quad t=\frac{\ln 2}{k}=\frac{\ln 2}{\tfrac{1}{2}\ln(1.1)}=\frac{2\ln 2}{\ln(1.1)}.$$
Numerically,
$$t=\frac{2\times 0.693147}{0.095310} \approx 14.55\ \text{hours}.$$
Answer
$$\boxed{t=\dfrac{2\ln 2}{\ln(1.1)}\approx 14.55\ \text{hours},}$$
Download concise revision notes by Anand Classes — perfect for JEE and CBSE practice and quick review of exponential growth models.
NCERT Question 23 : The general solution of the differential equation is
$$\frac{dy}{dx} = e^{x + y}$$
(A) $e^x + e^{-y} = C$
(B) $e^x + e^y = C$
(C) $e^{-x} + e^y = C$
(D) $e^{-x} + e^{-y} = C$
Solution :
Given,
$$\frac{dy}{dx} = e^{x + y} = e^x \cdot e^y$$
Separating the variables,
$$e^{-y}dy = e^xdx$$
Integrating both sides,
$$\int e^{-y}dy = \int e^xdx$$
We get,
$$-e^{-y} = e^x + C$$
Rewriting,
$$e^x + e^{-y} = -C$$
Letting $-C = C$, we have the general solution
$$\boxed{e^x + e^{-y} = C}$$
Hence, the correct option is (A).
For more conceptual explanations and solved examples, explore premium differential equation notes by Anand Classes, ideal for JEE and CBSE preparation.
Summary
Chapter 9 of the Class 12 NCERT Mathematics Part II textbook, “Differential Equations,” focuses on techniques for solving differential equations. Exercise 9.3 provides practice problems on applying methods for solving both first-order and higher-order differential equations. Solutions are detailed to help students understand and effectively apply these techniques to various problems.
