NCERT Class 12 Mathematics — Differential Equations, Exercise 9.2. All 12 questions solved.

Ten verifications and two definitions. A verification is always the same three steps:

1. Differentiate the given function as many times as the equation needs.
2. Substitute the results into the left-hand side of the differential equation.
3. Simplify and show it equals the right-hand side.

Key insight. For an implicit function — questions 7, 8, 9 and 10 — do not try to solve for $y$ first. Differentiate the relation as it stands, using the chain rule on every $y$, then rearrange for $\tfrac{dy}{dx}$. Solving for $y$ explicitly is usually impossible and never necessary.

The other habit worth forming: substitute into the left-hand side alone and simplify until it becomes the right-hand side. Manipulating both sides at once is how sign errors slip through unnoticed.

In each of the Exercises 1 to 10 verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

Question 1

$y = e^x + 1$  :  $y” – y’ = 0$

Solution. Differentiating twice:

$$y’ = e^x, \qquad y” = e^x$$

Substituting into the left-hand side:

$$y” – y’ = e^x – e^x = 0$$

which is the right-hand side. Hence $y = e^x + 1$ is a solution.

Question 2

$y = x^2 + 2x + \mathrm{C}$  :  $y’ – 2x – 2 = 0$

Solution. $y’ = 2x + 2$, so

$$y’ – 2x – 2 = (2x+2) – 2x – 2 = 0$$

The arbitrary constant $\mathrm{C}$ disappears on differentiation, which is why a whole family of curves satisfies the same equation.

Question 3

$y = \cos x + \mathrm{C}$  :  $y’ + \sin x = 0$

Solution. $y’ = -\sin x$, so $y’ + \sin x = -\sin x + \sin x = 0$.

Question 4

$y = \sqrt{1 + x^2}$  :  $y’ = \dfrac{xy}{1 + x^2}$

Solution. By the chain rule,

$$y’ = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$$

Now compare with the right-hand side, substituting $y = \sqrt{1+x^2}$:

$$\frac{xy}{1+x^2} = \frac{x\sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$$

The two agree, so the function is a solution.

Question 5

$y = \mathrm{A}x$  :  $xy’ = y$  $(x \ne 0)$

Solution. $y’ = \mathrm{A}$, so $xy’ = \mathrm{A}x = y$.

Question 6

$y = x\sin x$  :  $xy’ = y + x\sqrt{x^2 – y^2}$  $(x \ne 0$ and $x > y$ or $x < -y)$

Solution. By the product rule, $y’ = \sin x + x\cos x$, so

$$xy’ = x\sin x + x^2\cos x = y + x^2\cos x$$

For the right-hand side, note that $x^2 – y^2 = x^2 – x^2\sin^2 x = x^2\cos^2 x$, so

$$x\sqrt{x^2 – y^2} = x \cdot |x\cos x| = x^2\cos x$$

taking the positive root, which is what the stated restriction on $x$ and $y$ guarantees. The two sides match.

Question 7

$xy = \log y + \mathrm{C}$  :  $y’ = \dfrac{y^2}{1 – xy}$  $(xy \ne 1)$

Solution. This is implicit, so differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

$$y + xy’ = \frac{y’}{y}$$

Multiply through by $y$ and collect the $y’$ terms:

$$y^2 + xyy’ = y’ \quad\Longrightarrow\quad y^2 = y'(1 – xy)$$

$$y’ = \frac{y^2}{1 – xy}$$

The condition $xy \ne 1$ is exactly what keeps the last division legal.

Question 8

$y – \cos y = x$  :  $(y\sin y + \cos y + x)y’ = y$

Solution. Differentiating implicitly:

$$y’ + y’\sin y = 1 \quad\Longrightarrow\quad y'(1 + \sin y) = 1 \quad\Longrightarrow\quad y’ = \frac{1}{1 + \sin y}$$

Now substitute into the left-hand side, replacing $x$ by $y – \cos y$ from the given relation:

$$(y\sin y + \cos y + x)\,y’ = \big(y\sin y + \cos y + y – \cos y\big)\frac{1}{1+\sin y}$$

$$= \frac{y\sin y + y}{1 + \sin y} = \frac{y(1 + \sin y)}{1 + \sin y} = y$$

Using the original relation to eliminate $x$ is the step that makes this work.

Question 9

$x + y = \tan^{-1}y$  :  $y^2y’ + y^2 + 1 = 0$

Solution. Differentiating implicitly:

$$1 + y’ = \frac{y’}{1 + y^2}$$

Multiply by $1 + y^2$:

$$1 + y^2 + y’\left(1 + y^2\right) = y’ \quad\Longrightarrow\quad 1 + y^2 = y’ – y’ – y^2y’ = -y^2y’$$

Rearranging gives $y^2y’ + y^2 + 1 = 0$, as required.

Question 10

$y = \sqrt{a^2 – x^2}$, $x \in (-a, a)$  :  $x + y\dfrac{dy}{dx} = 0$  $(y \ne 0)$

Solution. Rather than differentiate the square root, square first: $y^2 = a^2 – x^2$. Differentiating that implicitly:

$$2y\frac{dy}{dx} = -2x \quad\Longrightarrow\quad y\frac{dy}{dx} = -x$$

so $x + y\tfrac{dy}{dx} = 0$. Squaring first turns a chain-rule calculation into one line.

Question 11

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

Solution. Each integration introduces one arbitrary constant, and solving an equation of order $n$ requires $n$ integrations. So the general solution of a fourth-order equation contains exactly $4$ arbitrary constants.

Question 12

The number of arbitrary constants in the particular solution of a differential equation of third order are:

Solution. A particular solution is one in which the arbitrary constants have already been determined from the given conditions. Whatever the order, a particular solution contains none.

The pair of questions 11 and 12 is the whole distinction: general solution keeps the constants, particular solution has pinned them down.

Common mistakes

• Trying to make an implicit function explicit. In question 7, solving $xy = \log y + \mathrm{C}$ for $y$ is impossible in elementary terms — and unnecessary.
• Forgetting the chain rule on $y$. Differentiating $\cos y$ gives $-y’\sin y$, not $-\sin y$. Every $y$ carries a $y’$.
• Not using the given relation to simplify. Question 8 only closes because $x$ is replaced by $y – \cos y$ partway through.
• Dropping the modulus on a square root. In question 6, $\sqrt{x^2\cos^2 x} = |x\cos x|$; the stated restriction is what lets you write it without bars.
• Confusing “general” with “particular”. Questions 11 and 12 differ only in that word, and the answers differ by the order of the equation.

Practise next

• Exercise 9.3 — actually solving differential equations by separating the variables, rather than verifying a given solution.
• Exercise 9.1 — the order and degree that questions 11 and 12 here refer to.