NCERT Exemplar Solutions Class 12 Maths Differential Equations — Objective Type Q55–75

NCERT Exemplar Problems, Class 12 Mathematics — Differential Equations, Objective Type questions 55 to 75. All 21 questions solved.

This is the second half of the objective band. Most questions are one-step recognitions of a method: separate the variables, or write the equation as $\dfrac{dy}{dx}+Py=Q$ and multiply by $e^{\int P\,dx}$. A few ask for order and degree, and a few ask you to eliminate constants from a family of curves. Work each one fully, then match the form to the option.

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 56, 59, 60, 65 and 69 are formation questions (outside the current textbook and Main, relevant to Advanced; Questions 56 and 65 can also be done textbook-style by checking each option, Ex 9.2 style). Question 70 involves a second-order equation, and its options are checked by substitution. All the others are fully in syllabus.

Key insight. Name the method before computing. A sum of $dx$ and $dy$ terms that splits into a function of $x$ times $dx$ and a function of $y$ times $dy$ is separable. An equation of the form $\dfrac{dy}{dx}+Py=Q$ is linear, and its integrating factor is $e^{\int P\,dx}$. A family with $n$ independent constants gives an equation of order $n$.

Question 55

The integrating factor of the differential equation $\dfrac{dy}{dx}+y=\dfrac{1+y}{x}$ is:

• (A) $\dfrac{x}{e^x}$
• (B) $\dfrac{e^x}{x}$
• (C) $xe^x$
• (D) $e^x$

Solution. This is linear, but $y$ appears on both sides, so first collect it: $\dfrac{dy}{dx}+\left(1-\dfrac1x\right)y=\dfrac1x$. Here $P=1-\dfrac1x$, so $\int P\,dx=x-\log x$ and the integrating factor is $e^{x-\log x}=\dfrac{e^x}{x}$. Reading off $P=1$ without moving the $\dfrac yx$ term across is the trap.

Question 56

$y=ae^{mx}+be^{-mx}$ satisfies which of the following differential equations?

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

Solution. This is a formation question: the family has two constants $a,b$, so the equation is of order two. Differentiating twice, $y’=m\left(ae^{mx}-be^{-mx}\right)$ and $y”=m^2\left(ae^{mx}+be^{-mx}\right)=m^2y$. Hence $\dfrac{d^2y}{dx^2}-m^2y=0$. Textbook-style, you can instead test the options: (A) and (B) are first order, and a two-constant family cannot satisfy them in general.

Question 57

The solution of the differential equation $\cos x\sin y\,dx+\sin x\cos y\,dy=0$ is:

• (A) $\dfrac{\sin x}{\sin y}=c$
• (B) $\sin x\sin y=c$
• (C) $\sin x+\sin y=c$
• (D) $\cos x\cos y=c$

Solution. Variables separable. Divide by $\sin x\sin y$ to get $\cot x\,dx+\cot y\,dy=0$. Integrating, $\log|\sin x|+\log|\sin y|=\log c$, so $\sin x\sin y=c$. Alternatively, the left side is exactly $d(\sin x\sin y)$.

Question 58

The solution of $x\dfrac{dy}{dx}+y=e^x$ is:

• (A) $y=\dfrac{e^x}{x}+\dfrac kx$
• (B) $y=xe^x+cx$
• (C) $y=xe^x+k$
• (D) $x=\dfrac{e^y}{y}+\dfrac ky$

Solution. Linear equation. The left side is the derivative of a product: $x\dfrac{dy}{dx}+y=\dfrac{d}{dx}(xy)$. So $\dfrac{d}{dx}(xy)=e^x$, giving $xy=e^x+k$, that is $y=\dfrac{e^x}{x}+\dfrac kx$. (Dividing by $x$ and using the integrating factor $x$ gives the same.)

Question 59

The differential equation of the family of curves $x^2+y^2-2ay=0$, where $a$ is an arbitrary constant, is:

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

Solution. Formation: eliminate $a$. From the family, $2a=\dfrac{x^2+y^2}{y}$. Differentiating, $2x+2y\dfrac{dy}{dx}-2a\dfrac{dy}{dx}=0$, so $x+y\dfrac{dy}{dx}=a\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2y}\dfrac{dy}{dx}$. Multiplying by $2y$: $2xy+2y^2\dfrac{dy}{dx}=(x^2+y^2)\dfrac{dy}{dx}$, which gives $(x^2-y^2)\dfrac{dy}{dx}=2xy$.

Question 60

The family $y=Ax+A^3$ of curves will correspond to a differential equation of order

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

Solution. The order of the equation of a family equals the number of independent arbitrary constants. Here there is only one, $A$, because $A^3$ is not a new constant. Eliminating $A$ needs one differentiation, $y’=A$, giving $y=xy’+(y’)^3$, an equation of order $1$ (and degree $3$).

Question 61

The general solution of $\dfrac{dy}{dx}=2x\,e^{x^2-y}$ is:

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

Solution. Variables separable. Write $e^{x^2-y}=e^{x^2}e^{-y}$, so $e^y\,dy=2x\,e^{x^2}dx$. Integrating, $e^y=e^{x^2}+c$, since $\dfrac{d}{dx}e^{x^2}=2xe^{x^2}$.

Question 62

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is:

• (A) an ellipse
• (B) parabola
• (C) circle
• (D) rectangular hyperbola

Solution. The slope is $\dfrac{dy}{dx}=\dfrac xy$. Separating, $y\,dy=x\,dx$, so $\dfrac{y^2}{2}=\dfrac{x^2}{2}+c$, i.e. $y^2-x^2=k$. This is a hyperbola with equal axes, a rectangular hyperbola.

Question 63

The general solution of the differential equation $\dfrac{dy}{dx}=e^{\frac{x^2}{2}}+xy$ is:

• (A) $y=ce^{\frac{-x^2}{2}}$
• (B) $y=ce^{\frac{x^2}{2}}$
• (C) $y=(x+c)e^{\frac{x^2}{2}}$
• (D) $y=(c-x)e^{\frac{x^2}{2}}$

Solution. Linear: $\dfrac{dy}{dx}-xy=e^{x^2/2}$, with $P=-x$ and integrating factor $e^{-x^2/2}$. Then $\dfrac{d}{dx}\left(ye^{-x^2/2}\right)=e^{x^2/2}\cdot e^{-x^2/2}=1$, so $ye^{-x^2/2}=x+c$ and $y=(x+c)e^{x^2/2}$.

Question 64

The solution of the equation $(2y-1)\,dx-(2x+3)\,dy=0$ is:

• (A) $\dfrac{2x-1}{2y+3}=k$
• (B) $\dfrac{2y+1}{2x-3}=k$
• (C) $\dfrac{2x+3}{2y-1}=k$
• (D) $\dfrac{2x-1}{2y-1}=k$

Solution. Variables separable: $\dfrac{dx}{2x+3}=\dfrac{dy}{2y-1}$. Integrating, $\tfrac12\log|2x+3|=\tfrac12\log|2y-1|+\tfrac12\log k$, so $\dfrac{2x+3}{2y-1}=k$.

Question 65

The differential equation for which $y=a\cos x+b\sin x$ is a solution, is:

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

Solution. Formation with two constants, so differentiate twice: $y’=-a\sin x+b\cos x$ and $y”=-a\cos x-b\sin x=-y$. Hence $y”+y=0$. Options (C) and (D) contain the constants $a,b$, which a differential equation of the family must not. Textbook-style: substituting $y=\cos x$ or $\sin x$ into each option leaves only (A) satisfied.

Question 66

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) $y=xe^{-x}$

Solution. Linear with integrating factor $e^x$: $\dfrac{d}{dx}(ye^x)=e^{-x}e^x=1$, so $ye^x=x+c$. The condition $y(0)=0$ gives $c=0$, hence $y=xe^{-x}$.

Question 67

The order and degree of the differential equation $\left(\dfrac{d^3y}{dx^3}\right)^2-3\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^4=y^4$ are:

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

Solution. The highest derivative is $\dfrac{d^3y}{dx^3}$, so the order is $3$. The equation is already a polynomial in the derivatives, and the highest derivative is raised to the power $2$, so the degree is $2$. The power $4$ of $\dfrac{dy}{dx}$ does not matter, since it is not the highest-order derivative.

Question 68

The order and degree of the differential equation $\left[1+\left(\dfrac{dy}{dx}\right)^2\right]=\dfrac{d^2y}{dx^2}$ are:

• (A) $2,\ \dfrac32$
• (B) $2,3$
• (C) $2,1$
• (D) $3,4$

Solution. The highest derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$. It appears to the first power, and the equation is a polynomial in the derivatives, so the degree is $1$. Had the bracket carried the power $\tfrac32$, we would square both sides first and the degree would change; here no fractional power appears.

Question 69

The differential equation of the family of curves $y^2=4a(x+a)$ is:

• (A) $y^2=4\dfrac{dy}{dx}\left(x+\dfrac{dy}{dx}\right)$
• (B) $2y\dfrac{dy}{dx}=4a$
• (C) $y\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$
• (D) $2x\dfrac{dy}{dx}+y\left(\dfrac{dy}{dx}\right)^2-y$

Solution. Formation: the family has one constant $a$, so the equation has order one. Differentiating, $2yy’=4a$, so $a=\dfrac{yy’}{2}$. Substituting back, $y^2=2yy’\left(x+\dfrac{yy’}{2}\right)=2xyy’+y^2y’^2$. Dividing by $y$: $y=2xy’+y\,(y’)^2$, i.e. $2x\dfrac{dy}{dx}+y\left(\dfrac{dy}{dx}\right)^2-y=0$. Option (D) is printed without the “$=0$”; we show it as printed and read it as an expression set equal to zero, which is the only reading that makes it an equation. Option (B) still contains the constant $a$, so it cannot be the equation of the family.

A note on the printed answer. The key prints (C). The equation $yy”+(y’)^2=0$ is indeed satisfied by every curve $y^2=4a(x+a)$, but it is second order. It is the equation of the larger two-parameter family $y^2=Ax+B$ (differentiate once to get $2yy’=A$, then again), of which our curves ($A=4a$, $B=4a^2$) are a special case. A family with one constant needs a first-order equation, so (D) is the correct choice.

Question 70

Which of the following is the general solution of $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$?

• (A) $y=(Ax+B)e^{x}$
• (B) $y=(Ax+B)e^{-x}$
• (C) $y=Ae^{x}+Be^{-x}$
• (D) $y=A\cos x+B\sin x$

Solution. This is a second-order equation, so we check the options by substitution (the general method is outside the current syllabus). The auxiliary equation is $m^2-2m+1=(m-1)^2=0$, with a repeated root $m=1$, which gives $y=(Ax+B)e^x$. Check: for $y=xe^x$, $y’=(x+1)e^x$ and $y”=(x+2)e^x$, so $y”-2y’+y=(x+2-2x-2+x)e^x=0$. Options (B), (C) and (D) involve roots $-1$ or $\pm i$ and fail.

Question 71

The general solution of $\dfrac{dy}{dx}+y\tan x=\sec x$ is:

• (A) $y\sec x=\tan x+c$
• (B) $y\tan x=\sec x+c$
• (C) $\tan x=y\tan x+c$
• (D) $x\sec x=\tan y+c$

Solution. Linear with $P=\tan x$, so the integrating factor is $e^{\int\tan x\,dx}=e^{\log\sec x}=\sec x$. Then $y\sec x=\int\sec^2x\,dx=\tan x+c$.

Question 72

The solution of the differential equation $\dfrac{dy}{dx}+\dfrac yx=\sin x$ is:

• (A) $x(y+\cos x)=\sin x+c$
• (B) $x(y-\cos x)=\sin x+c$
• (C) $xy\cos x=\sin x+c$
• (D) $x(y+\cos x)=\cos x+c$

Solution. Linear with $P=\dfrac1x$, integrating factor $x$. Then $xy=\int x\sin x\,dx=-x\cos x+\sin x+c$ (by parts). Moving the $x\cos x$ term across, $x(y+\cos x)=\sin x+c$.

Question 73

The general solution of the differential equation $(e^x+1)\,y\,dy=(y+1)\,e^x\,dx$ is:

• (A) $(y+1)=k(e^x+1)$
• (B) $y+1=e^x+1+k$
• (C) $y=\log\{k(y+1)(e^x+1)\}$
• (D) $y=\log\left\{\dfrac{e^x+1}{y+1}\right\}+k$

Solution. Variables separable: $\dfrac{y}{y+1}\,dy=\dfrac{e^x}{e^x+1}\,dx$. Write $\dfrac{y}{y+1}=1-\dfrac1{y+1}$, so $y-\log|y+1|=\log(e^x+1)+\log k$. Therefore $y=\log\{k(y+1)(e^x+1)\}$.

Question 74

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

• (A) $y=e^{x-y}-x^2e^{-y}+c$
• (B) $e^y-e^x=\dfrac{x^3}{3}+c$
• (C) $e^x+e^y=\dfrac{x^3}{3}+c$
• (D) $e^x-e^y=\dfrac{x^3}{3}+c$

Solution. Variables separable. Factor out $e^{-y}$: $\dfrac{dy}{dx}=e^{-y}\left(e^x+x^2\right)$, so $e^y\,dy=\left(e^x+x^2\right)dx$. Integrating, $e^y=e^x+\dfrac{x^3}{3}+c$, i.e. $e^y-e^x=\dfrac{x^3}{3}+c$.

Question 75

The solution of the differential equation $\dfrac{dy}{dx}+\dfrac{2xy}{1+x^2}=\dfrac1{(1+x^2)^2}$ is:

• (A) $y(1+x^2)=c+\tan^{-1}x$
• (B) $\dfrac{y}{1+x^2}=c+\tan^{-1}x$
• (C) $y\log(1+x^2)=c+\tan^{-1}x$
• (D) $y(1+x^2)=c+\sin^{-1}x$

Solution. Linear with $P=\dfrac{2x}{1+x^2}$, so the integrating factor is $e^{\log(1+x^2)}=1+x^2$. Then $y(1+x^2)=\int\dfrac{1}{(1+x^2)^2}(1+x^2)\,dx=\int\dfrac{dx}{1+x^2}=\tan^{-1}x+c$.

Common mistakes

• Reading the integrating factor before the equation is in standard form. In Question 55 the $y$ on the right side of $\dfrac{1+y}{x}$ must be moved across, which changes $P$ from $1$ to $1-\dfrac1x$.
• Counting the constants wrongly. In Question 60, $A$ and $A^3$ are one constant, so the order is $1$, not $3$ or $2$.
• Choosing a differential equation that the family satisfies but is of too high an order. In Question 69, $yy”+(y’)^2=0$ is satisfied by the family, but it describes the two-constant family $y^2=Ax+B$. The equation of a one-constant family is first order.
• Misreading the degree. In Question 67 the degree comes from the power of $\dfrac{d^3y}{dx^3}$ alone, not from the power $4$ on $\dfrac{dy}{dx}$.
• Losing a sign when integrating by parts or splitting a fraction. Question 72 needs $\int x\sin x\,dx=-x\cos x+\sin x$, and Question 73 needs $\dfrac{y}{y+1}=1-\dfrac1{y+1}$.

Practise next

• The other bands of this unit: Differential Equations, short answer (Q1-24), long answer (Q25-33), objective, first half (Q34-54) 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.