NCERT Class 12 Mathematics — Differential Equations, Exercise 9.3. All 23 questions solved.

Every question in this exercise is solved by separating the variables: get all the $y$’s and $dy$ on one side, all the $x$’s and $dx$ on the other, and integrate both sides.

$$\frac{dy}{dx} = g(x)h(y) \quad\Longrightarrow\quad \int\frac{dy}{h(y)} = \int g(x)\,dx$$

The exercise then splits into three parts: general solutions (1 to 10), particular solutions where a given condition fixes the constant (11 to 14), and word problems (15 to 22) where you must build the equation before solving it.

Key insight. Only one arbitrary constant is needed, not one per side. Write $+\,\mathrm{C}$ on the right and nothing on the left; and where both sides give logarithms, replace $\mathrm{C}$ by $\log\mathrm{C}$ at the outset so the whole equation collapses into a product rather than a sum. Question 7 and question 10 both come out far cleaner that way.

For the word problems, the phrase “rate of change” or “proportional to” is always the instruction to write $\tfrac{dy}{dt} = k\,(\cdots)$. Questions 20, 21 and 22 are three variations on the same exponential-growth equation.

For each of the differential equations in Exercises 1 to 10, find the general solution.

Question 1

$\dfrac{dy}{dx} = \dfrac{1 – \cos x}{1 + \cos x}$

Solution. Convert to half angles before integrating:

$$\frac{1-\cos x}{1+\cos x} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2} = \sec^2\frac{x}{2} – 1$$

$$y = \int\left(\sec^2\frac{x}{2} – 1\right)dx = 2\tan\frac{x}{2} – x + \mathrm{C}$$

Question 2

$\dfrac{dy}{dx} = \sqrt{4 – y^2}$  $(-2 < y < 2)$

Solution. Separate and integrate:

$$\int\frac{dy}{\sqrt{4 – y^2}} = \int dx \quad\Longrightarrow\quad \sin^{-1}\frac{y}{2} = x + \mathrm{C}$$

$$y = 2\sin(x + \mathrm{C})$$

Question 3

$\dfrac{dy}{dx} + y = 1$  $(y \ne 1)$

Solution. $\dfrac{dy}{1-y} = dx$, so

$$-\log|1-y| = x + c \quad\Longrightarrow\quad |1-y| = e^{-x-c}$$

$$y = 1 + \mathrm{A}e^{-x}$$

where $\mathrm{A} = \pm e^{-c}$ absorbs both the sign and the constant.

Question 4

$\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$

Solution. Divide throughout by $\tan x\tan y$:

$$\frac{\sec^2x}{\tan x}dx + \frac{\sec^2y}{\tan y}dy = 0$$

Each side is now of the form $\tfrac{f’}{f}$, so both integrate to logarithms:

$$\log|\tan x| + \log|\tan y| = \log \mathrm{C}$$

Question 5

$\left(e^x + e^{-x}\right)dy – \left(e^x – e^{-x}\right)dx = 0$

Solution. The numerator on the right is the derivative of the bracket on the left:

$$dy = \frac{e^x – e^{-x}}{e^x + e^{-x}}\,dx \quad\Longrightarrow\quad y = \log\left(e^x + e^{-x}\right) + \mathrm{C}$$

Question 6

$\dfrac{dy}{dx} = \left(1 + x^2\right)\left(1 + y^2\right)$

Solution. Already factored into a function of $x$ times a function of $y$:

$$\int\frac{dy}{1+y^2} = \int\left(1 + x^2\right)dx$$

Question 7

$y\log y\,dx – x\,dy = 0$

Solution. Separate:

$$\frac{dy}{y\log y} = \frac{dx}{x}$$

For the left side, put $t = \log y$ so $dt = \tfrac{dy}{y}$, giving $\int\tfrac{dt}{t} = \log|\log y|$:

$$\log|\log y| = \log|x| + \log c \quad\Longrightarrow\quad \log y = cx$$

Question 8

$x^5\dfrac{dy}{dx} = -y^5$

Solution.

$$\frac{dy}{y^5} = -\frac{dx}{x^5} \quad\Longrightarrow\quad \frac{y^{-4}}{-4} = \frac{x^{-4}}{4} + c$$

Multiplying by $-4$ and renaming the constant:

Question 9

$\dfrac{dy}{dx} = \sin^{-1}x$

Solution. Integrate the right side by parts with $1$ as the second function:

$$y = \int\sin^{-1}x\,dx = x\sin^{-1}x – \int\frac{x}{\sqrt{1-x^2}}dx = x\sin^{-1}x + \sqrt{1-x^2} + \mathrm{C}$$

Question 10

$e^x\tan y\,dx + \left(1 – e^x\right)\sec^2y\,dy = 0$

Solution. Separate:

$$\frac{\sec^2y}{\tan y}dy = \frac{-e^x}{1 – e^x}dx$$

The right-hand side is $\tfrac{f’}{f}$ with $f = 1 – e^x$, so both sides are logarithms:

$$\log|\tan y| = \log|1 – e^x| + \log \mathrm{C}$$

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition.

Question 11

$\left(x^3 + x^2 + x + 1\right)\dfrac{dy}{dx} = 2x^2 + x$; $y = 1$ when $x = 0$

Solution. Factor the cubic by grouping:

$$x^3 + x^2 + x + 1 = x^2(x+1) + (x+1) = (x+1)\left(x^2+1\right)$$

Now use partial fractions on $\dfrac{2x^2+x}{(x+1)(x^2+1)}$. Writing it as $\dfrac{A}{x+1} + \dfrac{Bx + C}{x^2+1}$ and comparing coefficients gives $A = \tfrac12$, $B = \tfrac32$, $C = -\tfrac12$:

$$y = \int\left[\frac{1}{2(x+1)} + \frac{3x – 1}{2\left(x^2+1\right)}\right]dx = \frac12\log|x+1| + \frac34\log\left(x^2+1\right) – \frac12\tan^{-1}x + c$$

At $x = 0$, $y = 1$ gives $c = 1$. Combining the two logarithms:

Question 12

$x\left(x^2 – 1\right)\dfrac{dy}{dx} = 1$; $y = 0$ when $x = 2$

Solution. Partial fractions on $\dfrac{1}{x(x-1)(x+1)}$ give $-\dfrac1x + \dfrac{1}{2(x-1)} + \dfrac{1}{2(x+1)}$, so

$$y = -\log|x| + \frac12\log|x-1| + \frac12\log|x+1| + c = \frac12\log\left|\frac{x^2-1}{x^2}\right| + c$$

At $x = 2$, $y = 0$: $0 = \tfrac12\log\tfrac34 + c$, so $c = -\tfrac12\log\tfrac34$.

Question 13

$\cos\left(\dfrac{dy}{dx}\right) = a$  $(a \in \mathbf{R})$; $y = 1$ when $x = 0$

Solution. Take the inverse cosine of both sides. Since $a$ is a constant, so is $\cos^{-1}a$:

$$\frac{dy}{dx} = \cos^{-1}a$$

The derivative is constant, so $y$ is linear:

$$y = x\cos^{-1}a + \mathrm{C}$$

Applying $y = 1$ at $x = 0$ gives $\mathrm{C} = 1$. Rearranging:

$$\frac{y – 1}{x} = \cos^{-1}a \quad\Longrightarrow\quad \cos\left(\frac{y-1}{x}\right) = a$$

A note on the printed answer. The NCERT key gives $\cos\left(\tfrac{y-2}{x}\right) = a$. That answer corresponds to the initial condition $y = 2$ at $x = 0$, which is how this question read in an earlier edition of the book. With the condition printed in the current edition — $y = 1$ when $x = 0$ — the constant is $1$ and the answer above is the correct one. If your work gave $\tfrac{y-1}{x}$, it was right.

Question 14

$\dfrac{dy}{dx} = y\tan x$; $y = 1$ when $x = 0$

Solution.

$$\frac{dy}{y} = \tan x\,dx \quad\Longrightarrow\quad \log|y| = \log|\sec x| + \log \mathrm{C}$$

so $y = \mathrm{C}\sec x$. At $x = 0$, $\sec 0 = 1$, so $\mathrm{C} = 1$.

Question 15

Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y’ = e^x\sin x$.

Solution. The integral $\int e^x\sin x\,dx$ is the classic integrate-by-parts-twice-and-solve-for-$I$ problem, giving

$$\int e^x\sin x\,dx = \frac{e^x(\sin x – \cos x)}{2}$$

So $y = \tfrac{e^x(\sin x – \cos x)}{2} + \mathrm{C}$. At $(0, 0)$: $0 = \tfrac{1(0-1)}{2} + \mathrm{C}$, so $\mathrm{C} = \tfrac12$.

Question 16

For the differential equation $xy\dfrac{dy}{dx} = (x+2)(y+2)$, find the solution curve passing through the point $(1, -1)$.

Solution. Separate:

$$\frac{y}{y+2}dy = \frac{x+2}{x}dx$$

Both sides are improper fractions, so divide first:

$$\int\left(1 – \frac{2}{y+2}\right)dy = \int\left(1 + \frac2x\right)dx$$

$$y – 2\log|y+2| = x + 2\log|x| + c$$

At $(1, -1)$: $-1 – 2\log 1 = 1 + 2\log 1 + c$, so $c = -2$. Rearranging and combining logarithms:

Question 17

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 of the point is equal to the $x$ coordinate of the point.

Solution. “Product of the slope and $y$ equals $x$” translates directly:

$$y\frac{dy}{dx} = x \quad\Longrightarrow\quad y\,dy = x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + c$$

At $(0, -2)$: $2 = 0 + c$, so $c = 2$.

Question 18

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. The slope of the segment from $(x, y)$ to $(-4, -3)$ is $\dfrac{y+3}{x+4}$, so the condition is

$$\frac{dy}{dx} = \frac{2(y+3)}{x+4}$$

Separating:

$$\frac{dy}{y+3} = \frac{2\,dx}{x+4} \quad\Longrightarrow\quad \log|y+3| = 2\log|x+4| + \log \mathrm{C}$$

$$y + 3 = \mathrm{C}(x+4)^2$$

At $(-2, 1)$: $4 = \mathrm{C}(4)$, so $\mathrm{C} = 1$.

Question 19

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 the balloon after $t$ seconds.

Solution. “Changes at a constant rate” means $\dfrac{dV}{dt} = k$, so $V = kt + c$ with $V = \tfrac43\pi r^3$.

At $t = 0$, $r = 3$: $V = 36\pi$, so $c = 36\pi$. At $t = 3$, $r = 6$: $V = 288\pi$, so $288\pi = 3k + 36\pi$ and $k = 84\pi$.

$$\frac43\pi r^3 = 84\pi t + 36\pi \quad\Longrightarrow\quad r^3 = 63t + 27$$

Question 20

In a bank, principal increases continuously at the rate of $r\%$ per year. Find the value of $r$ if ₹100 double itself in $10$ years $(\log_e 2 = 0.6931)$.

Solution. Continuous growth means

$$\frac{dP}{dt} = \frac{r}{100}P \quad\Longrightarrow\quad P = P_0e^{rt/100}$$

Doubling in $10$ years gives $2 = e^{10r/100} = e^{r/10}$, so

$$\frac{r}{10} = \log_e 2 = 0.6931 \quad\Longrightarrow\quad r = 6.931$$

Question 21

In a bank, principal increases continuously at the rate of $5\%$ per year. An amount of ₹1000 is deposited with this bank, how much will it be worth after $10$ years $\left(e^{0.5} = 1.648\right)$?

Solution. Same model with $r = 5$:

$$P = 1000\,e^{0.05 \times 10} = 1000\,e^{0.5} = 1000(1.648)$$

Question 22

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. “Proportional to the number present” gives $\dfrac{dN}{dt} = kN$, so $N = N_0e^{kt}$ with $N_0 = 100000$.

From $N(2) = 110000$:

$$e^{2k} = \frac{11}{10} \quad\Longrightarrow\quad k = \frac12\log\frac{11}{10}$$

We want $N(t) = 200000$, that is $e^{kt} = 2$:

$$t = \frac{\log 2}{k} = \frac{2\log 2}{\log\frac{11}{10}}$$

Question 23

The general solution of the differential equation $\dfrac{dy}{dx} = e^{x+y}$ is

Solution. Split the exponential — this is the step that makes the variables separable at all:

$$\frac{dy}{dx} = e^x \cdot e^y \quad\Longrightarrow\quad e^{-y}dy = e^x dx$$

$$-e^{-y} = e^x + c \quad\Longrightarrow\quad e^x + e^{-y} = \mathrm{C}$$

Common mistakes

• Writing a constant on both sides. One arbitrary constant is enough; two merge into one anyway, and carrying both makes the final simplification harder.
• Not writing the constant as $\log \mathrm{C}$ when both sides are logarithms. In questions 4, 7, 10 and 18 that single choice turns a sum of logarithms into a clean product.
• Integrating an improper fraction without dividing. Question 16 needs $\tfrac{y}{y+2} = 1 – \tfrac{2}{y+2}$ before it can be integrated.
• Applying the initial condition before simplifying. In question 11 it is much easier to combine the logarithms first and then substitute.
• Forgetting to convert $r\%$ into $\tfrac{r}{100}$. Questions 20 and 21 both turn on it.
• Leaving the answer as $\log$ of something when the key wants it exponentiated, or the reverse. Both forms are correct; match the key’s form when checking.

Practise next

• Exercise 9.4 — homogeneous equations, where the substitution $y = vx$ is what makes the variables separable.
• Exercise 9.5 — linear equations solved with an integrating factor.