NCERT Exemplar Solutions Class 12 Maths Differential Equations — Objective Type Q34–54

NCERT Exemplar Problems, Class 12 Mathematics — Differential Equations, Objective Type questions 34 to 54. All 21 questions solved.

These are the quick-fire questions of the unit. Most ask for the order or degree of an equation, the integrating factor of a linear equation, or the solution of a separable or linear equation, so one standard step settles each. Where the answer is an option letter, the working shows why the other options fail.

A note on the syllabus. The rationalised NCERT Chapter 9 keeps order and degree (Ex 9.1), general and particular solutions (Ex 9.2), and solving first-order first-degree equations: variables separable (9.3), homogeneous (9.4) and linear (9.5). It dropped formation of a differential equation from a given family of curves. JEE Main 2026 lists order and degree, separation of variables, homogeneous and linear equations, and omits formation. JEE Advanced 2026 lists “formation of ordinary differential equations”. On this page, Questions 38, 42, 48 and 49 involve formation (outside the current textbook and Main, relevant to Advanced), and Question 37 is a second-order equation checked by substitution. Degree is read here as in JEE: remove radicals and fractional powers of derivatives first, then take the power of the highest-order derivative; the degree is not defined only when a derivative sits inside a transcendental function. The exemplar key uses a stricter reading, which matters in Questions 35 and 36.

Key insight. Put every equation in standard form before reading anything off it. For order and degree, clear fractional powers and look only at the highest derivative; for a linear equation, divide until the coefficient of $\dfrac{dy}{dx}$ is $1$ and the integrating factor is $e^{\int P\,dx}$.

Question 34

The degree of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^2=x\sin\left(\dfrac{dy}{dx}\right)$ is:

• (A) 1
• (B) 2
• (C) 3
• (D) not defined

Solution. The right-hand side contains $\dfrac{dy}{dx}$ inside the transcendental function $\sin$. The equation therefore cannot be written as a polynomial in the derivatives, and the degree is not defined (the squares on the left do not rescue it).

Question 35

The degree of the differential equation $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}=\dfrac{d^2y}{dx^2}$ is

• (A) $4$
• (B) $\dfrac32$
• (C) not defined
• (D) $2$

Solution. Remove the fractional power by squaring both sides: $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^3=\left(\dfrac{d^2y}{dx^2}\right)^2$. The highest-order derivative is $\dfrac{d^2y}{dx^2}$ and it appears to the power $2$, so the degree is $2$ (the left side expands to a polynomial in $\frac{dy}{dx}$, which is allowed).

A note on the printed answer. The key prints (C), not defined. It uses a stricter reading in which any fractional power makes the degree undefined. After rationalising, as JEE and most Indian textbooks do, the equation is a polynomial in the derivatives with degree $2$.

Question 36

The order and degree of the differential equation $\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^{\frac14}+x^{\frac15}=0$, respectively, are

• (A) $2$ and not defined
• (B) $2$ and $2$
• (C) $2$ and $3$
• (D) $3$ and $3$

Solution. The highest derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$To clear the fractional power, write $\dfrac{d^2y}{dx^2}+x^{1/5}=-\left(\dfrac{dy}{dx}\right)^{1/4}$ and raise to the fourth power: $\left(\dfrac{d^2y}{dx^2}+x^{1/5}\right)^4=\dfrac{dy}{dx}$. Read literally, the degree is then $4$, which is not among the options. The exemplar adopts the stricter convention in which a fractional power of a derivative makes the degree not defined, and with that reading the pair is $2$ and not defined, which is option (A). So (A) is the intended answer.

Question 37

If $y=e^{-x}(A\cos x+B\sin x)$, then $y$ is a solution of

• (A) $\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}=0$
• (B) $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+2y=0$
• (C) $\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}+2y=0$
• (D) $\dfrac{d^2y}{dx^2}+2y=0$

Solution. This is a substitution check. The function has the form $e^{mx}(A\cos x+B\sin x)$ with $m=-1$ and frequency $1$, which corresponds to characteristic roots $-1\pm i$, i.e. $r^2+2r+2=0$. To confirm directly, write $u=e^{-x}(-A\sin x+B\cos x)$. Then $y’=-y+u$ and $u’=-u-y$, so $y”=-y’+u’=-y’-u-y=-y’-(y’+y)-y=-2y’-2y$. So $y”+2y’+2y=0$.

Question 38

The differential equation for $y=A\cos\alpha x+B\sin\alpha x$, where $A$ and $B$ are arbitrary constants, is

• (A) $\dfrac{d^2y}{dx^2}-\alpha^2y=0$
• (B) $\dfrac{d^2y}{dx^2}+\alpha^2y=0$
• (C) $\dfrac{d^2y}{dx^2}+\alpha y=0$
• (D) $\dfrac{d^2y}{dx^2}-\alpha y=0$

Solution. Formation: two arbitrary constants need two differentiations. $y’=-A\alpha\sin\alpha x+B\alpha\cos\alpha x$ and $y”=-A\alpha^2\cos\alpha x-B\alpha^2\sin\alpha x=-\alpha^2y$. Hence $\dfrac{d^2y}{dx^2}+\alpha^2y=0$, which has no $A$ or $B$ left. Checking the options instead (as in Exercise 9.2) gives the same result.

Question 39

Solution of the differential equation $x\,dy-y\,dx=0$ represents:

• (A) a rectangular hyperbola
• (B) parabola whose vertex is at origin
• (C) straight line passing through origin
• (D) a circle whose centre is at origin

Solution. Variables separable: $\dfrac{dy}{y}=\dfrac{dx}{x}$, so $\log|y|=\log|x|+\log|c|$, i.e. $y=cx$. This is a family of straight lines through the origin.

Question 40

Integrating factor of the differential equation $\cos x\dfrac{dy}{dx}+y\sin x=1$ is:

• (A) $\cos x$
• (B) $\tan x$
• (C) $\sec x$
• (D) $\sin x$

Solution. Linear equation: first divide by $\cos x$ to reach the standard form $\dfrac{dy}{dx}+y\tan x=\sec x$. Then $\text{IF}=e^{\int\tan x\,dx}=e^{\log|\sec x|}=\sec x$.

Question 41

Solution of the differential equation $\tan y\sec^2x\,dx+\tan x\sec^2y\,dy=0$ is:

• (A) $\tan x+\tan y=k$
• (B) $\tan x-\tan y=k$
• (C) $\dfrac{\tan x}{\tan y}=k$
• (D) $\tan x\cdot\tan y=k$

Solution. Variables separable: divide by $\tan x\tan y$ to get $\dfrac{\sec^2x}{\tan x}dx+\dfrac{\sec^2y}{\tan y}dy=0$. Each term is a logarithmic derivative, so $\log|\tan x|+\log|\tan y|=\log k$, i.e. $\tan x\tan y=k$.

Question 42

Family $y=Ax+A^3$ of curves is represented by the differential equation of degree:

• (A) $1$
• (B) $2$
• (C) $3$
• (D) $4$

Solution. Formation: differentiating gives $\dfrac{dy}{dx}=A$. Substituting $A=\dfrac{dy}{dx}$ back, $y=x\dfrac{dy}{dx}+\left(\dfrac{dy}{dx}\right)^3$. The highest-order derivative is $\dfrac{dy}{dx}$, to the power $3$, so the degree is $3$.

A note on the printed answer. The key prints (A). The cubic term $\left(\frac{dy}{dx}\right)^3$ cannot be removed, so the degree is $3$, not $1$.

Question 43

Integrating factor of $x\dfrac{dy}{dx}-y=x^4-3x$ is:

• (A) $x$
• (B) $\log x$
• (C) $\dfrac1x$
• (D) $-x$

Solution. Linear equation: divide by $x$ to get $\dfrac{dy}{dx}-\dfrac1x\,y=x^3-3$. Here $P=-\dfrac1x$, so $\text{IF}=e^{-\int dx/x}=e^{-\log x}=\dfrac1x$.

Question 44

Solution of $\dfrac{dy}{dx}-y=1$, $y(0)=1$ is given by

• (A) $xy=-e^x$
• (B) $xy=-e^{-x}$
• (C) $xy=-1$
• (D) $y=2e^x-1$

Solution. Separate: $\dfrac{dy}{1+y}=dx$, so $\log|1+y|=x+c$, i.e. $1+y=Ce^x$. The condition $y(0)=1$ gives $C=2$, so $y=2e^x-1$. (Linear method with IF $e^{-x}$ gives the same.)

Question 45

The number of solutions of $\dfrac{dy}{dx}=\dfrac{y+1}{x-1}$ when $y(1)=2$ is:

• (A) none
• (B) one
• (C) two
• (D) infinite

Solution. Separate: $\dfrac{dy}{y+1}=\dfrac{dx}{x-1}$, giving $y+1=C(x-1)$. Every member of this family passes through $(1,-1)$, so at $x=1$ we always have $y=-1$. No choice of $C$ gives $y(1)=2$, so there is no solution.

A note on the printed answer. The key prints (B), one solution. The family $y+1=C(x-1)$ forces $y(1)=-1$ for every $C$, so the condition $y(1)=2$ can never be met.

Question 46

Which of the following is a second order differential equation?

• (A) $(y’)^2+x=y^2$
• (B) $y’y’+y=\sin x$
• (C) $y”+(y’)^2+y=0$
• (D) $y’=y^2$

Solution. The order is the order of the highest derivative present. Options (A) and (D) contain only $y’$. Option (B), as printed, is $y’\,y’+y=\sin x$ with two single primes, which is $(y’)^2+y=\sin x$, so it is first order too. Only (C) contains $y”$, so only (C) is second order.

A note on the printed answer. The key prints (B). As printed, option (B) has only first derivatives, so it is first order. The key would be right only if (B) were $y”y’+y=\sin x$, which is not what the question says.

Question 47

Integrating factor of the differential equation $(1-x^2)\dfrac{dy}{dx}-xy=1$ is

• (A) $-x$
• (B) $\dfrac{x}{1+x^2}$
• (C) $\sqrt{1-x^2}$
• (D) $\dfrac12\log(1-x^2)$

Solution. Linear: divide by $1-x^2$ to get $\dfrac{dy}{dx}-\dfrac{x}{1-x^2}\,y=\dfrac1{1-x^2}$. Then $\int P\,dx=-\int\dfrac{x}{1-x^2}dx=\dfrac12\log(1-x^2)$, so $\text{IF}=e^{\frac12\log(1-x^2)}=\sqrt{1-x^2}$. Option (D) is the exponent, not the factor.

Question 48

$\tan^{-1}x+\tan^{-1}y=c$ is the general solution of the differential equation:

• (A) $\dfrac{dy}{dx}=\dfrac{1+y^2}{1+x^2}$
• (B) $\dfrac{dy}{dx}=\dfrac{1+x^2}{1+y^2}$
• (C) $(1+x^2)\,dy+(1+y^2)\,dx=0$
• (D) $(1+x^2)\,dx+(1+y^2)\,dy=0$

Solution. Formation: differentiate $\tan^{-1}x+\tan^{-1}y=c$ to eliminate $c$: $\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\dfrac{dy}{dx}=0$. Multiplying by $(1+x^2)(1+y^2)$ gives $(1+y^2)\,dx+(1+x^2)\,dy=0$, which is option (C). Options (A) and (B) have the wrong sign.

Question 49

The differential equation $y\dfrac{dy}{dx}+x=c$ represents:

• (A) Family of hyperbolas
• (B) Family of parabolas
• (C) Family of ellipses
• (D) Family of circles

Solution. Variables separable: $y\,dy=(c-x)\,dx$. Integrating, $\dfrac{y^2}2=cx-\dfrac{x^2}2+d$, so $x^2+y^2=2cx+d$. Completing the square, $(x-c)^2+y^2=c^2+d$, a family of circles with centres on the $x$-axis.

Question 50

The general solution of $e^x\cos y\,dx-e^x\sin y\,dy=0$ is:

• (A) $e^x\cos y=k$
• (B) $e^x\sin y=k$
• (C) $e^x=k\cos y$
• (D) $e^x=k\sin y$

Solution. Divide by $e^x\cos y$: $dx-\tan y\,dy=0$, so $x+\log|\cos y|=\log k$, i.e. $e^x\cos y=k$. This is also seen directly, since the left side is $d(e^x\cos y)$.

Question 51

The degree of the differential equation $\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^3+6y^5=0$ is:

• (A) $1$
• (B) $2$
• (C) $3$
• (D) $5$

Solution. The highest-order derivative is $\dfrac{d^2y}{dx^2}$, and it appears to the first power. The powers $3$ and $5$ belong to a lower derivative and to $y$ itself, so they do not matter. The degree is $1$.

Question 52

The solution of $\dfrac{dy}{dx}+y=e^{-x}$, $y(0)=0$ is:

• (A) $y=e^x(x-1)$
• (B) $y=xe^{-x}$
• (C) $y=xe^{-x}+1$
• (D) $y=(x+1)e^{-x}$

Solution. Linear with $P=1$, $\text{IF}=e^x$. Then $\dfrac{d}{dx}(ye^x)=e^{-x}e^x=1$, so $ye^x=x+C$. With $y(0)=0$ we get $C=0$, and $y=xe^{-x}$.

Question 53

Integrating factor of the differential equation $\dfrac{dy}{dx}+y\tan x-\sec x=0$ is:

• (A) $\cos x$
• (B) $\sec x$
• (C) $e^{\cos x}$
• (D) $e^{\sec x}$

Solution. In standard form $\dfrac{dy}{dx}+(\tan x)\,y=\sec x$, so $P=\tan x$ and $\text{IF}=e^{\int\tan x\,dx}=e^{\log|\sec x|}=\sec x$.

Question 54

The solution of the differential equation $\dfrac{dy}{dx}=\dfrac{1+y^2}{1+x^2}$ is:

• (A) $y=\tan^{-1}x$
• (B) $y-x=k(1+xy)$
• (C) $x=\tan^{-1}y$
• (D) $\tan(xy)=k$

Solution. Variables separable: $\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}$, so $\tan^{-1}y-\tan^{-1}x=C$. Taking the tangent of both sides and using $\tan(a-b)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ gives $\dfrac{y-x}{1+xy}=\tan C=k$, i.e. $y-x=k(1+xy)$.

Common mistakes

• Reading the degree before rationalising. In Question 35 the power $\frac32$ looks like the degree; it is removed by squaring, giving degree $2$. Only the power of the highest-order derivative counts, and only after radicals are cleared.
• Counting powers of lower derivatives. In Question 51 the cubes and fifth powers belong to $\frac{dy}{dx}$ and $y$, not to $\frac{d^2y}{dx^2}$, so the degree is $1$.
• Forgetting to divide before finding the integrating factor. Questions 40, 43 and 47 have a coefficient other than $1$ in front of $\frac{dy}{dx}$. Skipping the division gives the wrong $P$, and so the wrong factor.
• Confusing the exponent with the factor. In Question 47, $\frac12\log(1-x^2)$ is $\int P\,dx$; the integrating factor is $e$ raised to it, namely $\sqrt{1-x^2}$.
• Assuming a solution always exists. In Question 45 every curve of the family $y+1=C(x-1)$ passes through $(1,-1)$, so the condition $y(1)=2$ cannot be satisfied and the answer is “none”.
• Misreading a misprinted option. In Question 46 option (B) has two single primes and is first order; only the option containing $y”$ is second order.

Practise next

• The other bands of this unit: short answer (Q1-24), long answer (Q25-33), objective, second half (Q55-75) and fill in the blanks and true or false (Q76-77).
• Textbook practice: Exercise 9.1, order and degree, Exercise 9.2, general and particular solutions, Exercise 9.3, variables separable, Exercise 9.5, linear equations and the Miscellaneous Exercise.