NCERT Exemplar Solutions Class 12 Maths Differential Equations — Short Answer Q1–24

NCERT Exemplar Problems, Class 12 Mathematics — Differential Equations, Exercise 9.3, Short Answer questions 1 to 24. All 24 questions solved.

These are the first-order workhorses of the unit. Most can be recognised in seconds: if the variables can be pulled to opposite sides it is variables separable; if every term has the same total degree it is homogeneous and $y=vx$ (or $x=vy$) works; if it has the shape $\dfrac{dy}{dx}+Py=Q$ it is linear and needs an integrating factor. A few need a substitution first, and a few go the other way, asking you to build the equation from a family of curves by eliminating constants.

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 2, 13, 14, 22 and 24 are formation questions, so they matter for JEE Advanced but not for the rationalised textbook or JEE Main. Every other question here is fully in syllabus.

Key insight. Classify before you compute. Separable, homogeneous, linear, or “needs a substitution such as $v=x+y$”: once the type is named, the rest is routine integration, and for formation questions the number of arbitrary constants fixes the order of the equation.

Question 1

Find the solution of $\dfrac{dy}{dx}=2^{y-x}$.

Solution. Method: variables separable. Write $2^{y-x}=\dfrac{2^y}{2^x}$, so $2^{-y}\,dy=2^{-x}\,dx$. Integrating both sides, $-\dfrac{2^{-y}}{\log 2}=-\dfrac{2^{-x}}{\log 2}+c$. Multiplying through by $-\log 2$ and absorbing the constant gives $2^{-x}-2^{-y}=k$, where $k$ is an arbitrary constant.

Question 2

Find the differential equation of all non vertical lines in a plane.

Solution. Method: formation. A non-vertical line has the form $y=mx+c$, with two arbitrary constants $m$ and $c$, so the equation will be of order two. Differentiating once gives $\dfrac{dy}{dx}=m$, and differentiating again gives $\dfrac{d^2y}{dx^2}=0$. Both constants have disappeared, so this is the required equation. It says the slope never changes, which is exactly what makes a line a line.

Question 3

Given that $\dfrac{dy}{dx}=e^{-2y}$ and $y=0$ when $x=5$. Find the value of $x$ when $y=3$.

Solution. Method: variables separable with a particular solution. Separate to get $e^{2y}\,dy=dx$, so $\dfrac{e^{2y}}{2}=x+c$. Use $y=0$ at $x=5$: $\dfrac12=5+c$, hence $c=-\dfrac92$. The solution is $e^{2y}=2x-9$. Now put $y=3$: $e^{6}=2x-9$, so $x=\dfrac{e^6+9}{2}$.

Question 4

Solve the differential equation $(x^2-1)\dfrac{dy}{dx}+2xy=\dfrac{1}{x^2-1}$.

Solution. Method: linear (integrating factor), though there is a shortcut. The left-hand side is exactly $\dfrac{d}{dx}\big[(x^2-1)y\big]$, because the derivative of $x^2-1$ is $2x$. So the equation reads $\dfrac{d}{dx}\big[(x^2-1)y\big]=\dfrac{1}{x^2-1}$. Integrating, and using $\displaystyle\int\frac{dx}{x^2-1}=\frac12\log\left|\frac{x-1}{x+1}\right|$, $$(x^2-1)\,y=\frac12\log\left|\frac{x-1}{x+1}\right|+C.$$ Dividing the standard form by $x^2-1$ gives integrating factor $e^{\int 2x/(x^2-1)\,dx}=x^2-1$, which is the same thing.

Question 5

Solve the differential equation $\dfrac{dy}{dx}+2xy=y$.

Solution. Method: variables separable (it is also linear, but separation is quicker). Move the $2xy$ across: $\dfrac{dy}{dx}=y(1-2x)$. Then $\dfrac{dy}{y}=(1-2x)\,dx$, so $\log|y|=x-x^2+c$, which gives $y=Ce^{x-x^2}$, with $C=\pm e^{c}$ an arbitrary constant.

Question 6

Find the general solution of $\dfrac{dy}{dx}+ay=e^{mx}$.

Solution. Method: linear. Here $P=a$ is constant, so the integrating factor is $e^{ax}$. Then $\dfrac{d}{dx}\big(ye^{ax}\big)=e^{(m+a)x}$. For $m\neq-a$, integrating gives $ye^{ax}=\dfrac{e^{(m+a)x}}{m+a}+C$, so $$y=\frac{e^{mx}}{m+a}+Ce^{-ax},\qquad\text{equivalently}\quad (a+m)\,y=e^{mx}+c\,e^{-ax}.$$ If $m=-a$ the right side becomes $1$ after multiplying by the integrating factor, and the solution is instead $y=(x+C)e^{-ax}$.

Question 7

Solve the differential equation $\dfrac{dy}{dx}+1=e^{x+y}$.

Solution. Method: substitution. The right side depends only on $x+y$, so put $v=x+y$. Then $\dfrac{dv}{dx}=1+\dfrac{dy}{dx}$, and the equation becomes $\dfrac{dv}{dx}=e^{v}$. Separating, $e^{-v}\,dv=dx$, so $-e^{-v}=x+c_1$, that is $e^{-(x+y)}=c-x$ with $c=-c_1$. Equivalently, $(x-c)\,e^{x+y}+1=0$.

Question 8

Solve: $y\,dx-x\,dy=x^2y\,dx$.

Solution. Method: variables separable. Divide the whole equation by $xy$ to get $\dfrac{dx}{x}-\dfrac{dy}{y}=x\,dx$. Integrating, $\log|x|-\log|y|=\dfrac{x^2}{2}+C$, so $\log\left|\dfrac xy\right|=\dfrac{x^2}{2}+C$. Solving for $y$, $y=kx\,e^{-x^2/2}$, where $k$ is an arbitrary non-zero constant.

Question 9

Solve the differential equation $\dfrac{dy}{dx}=1+x+y^2+xy^2$, when $y=0$, $x=0$.

Solution. Method: variables separable. Factorise by grouping: $1+x+y^2(1+x)=(1+x)(1+y^2)$. So $\dfrac{dy}{1+y^2}=(1+x)\,dx$, and integrating, $\tan^{-1}y=x+\dfrac{x^2}{2}+c$. Since $y=0$ when $x=0$, $c=0$. Hence $y=\tan\left(x+\dfrac{x^2}{2}\right)$.

Question 10

Find the general solution of $(x+2y^3)\dfrac{dy}{dx}=y$.

Solution. Method: linear, with $x$ as the dependent variable. As written, the equation is not linear in $y$, but flipping it gives $\dfrac{dx}{dy}=\dfrac{x+2y^3}{y}$, that is $\dfrac{dx}{dy}-\dfrac{x}{y}=2y^2$. This is linear in $x$ with $P=-\dfrac1y$, so the integrating factor is $e^{-\log y}=\dfrac1y$. Then $\dfrac{d}{dy}\left(\dfrac xy\right)=2y$, so $\dfrac xy=y^2+c$, and $x=y(y^2+c)$.

Question 11

If $y(x)$ is a solution of $\left(\dfrac{2+\sin x}{1+y}\right)\dfrac{dy}{dx}=-\cos x$ and $y(0)=1$, then find the value of $y\left(\dfrac{\pi}{2}\right)$.

Solution. Method: variables separable. Separate to get $\dfrac{dy}{1+y}=-\dfrac{\cos x}{2+\sin x}\,dx$. The right side is the derivative of $-\log(2+\sin x)$, so $\log|1+y|=-\log(2+\sin x)+c$, which gives $(1+y)(2+\sin x)=k$. At $x=0$, $y=1$: $2\cdot2=4=k$. At $x=\dfrac\pi2$: $(1+y)\cdot3=4$, so $y=\dfrac13$.

Question 12

If $y(t)$ is a solution of $(1+t)\dfrac{dy}{dt}-ty=1$ and $y(0)=-1$, then show that $y(1)=-\dfrac12$.

Solution. Method: linear. Divide by $1+t$: $\dfrac{dy}{dt}-\dfrac{t}{1+t}\,y=\dfrac{1}{1+t}$. Since $\dfrac{t}{1+t}=1-\dfrac1{1+t}$, $\displaystyle\int P\,dt=-t+\log(1+t)$, so the integrating factor is $e^{-t+\log(1+t)}=(1+t)e^{-t}$. Then $$\frac{d}{dt}\big[y(1+t)e^{-t}\big]=\frac{1}{1+t}\cdot(1+t)e^{-t}=e^{-t},$$ so $y(1+t)e^{-t}=-e^{-t}+C$. Using $y(0)=-1$: $-1=-1+C$, hence $C=0$ and $y=-\dfrac{1}{1+t}$. Therefore $y(1)=-\dfrac12$, as required.

Question 13

Form the differential equation having $y=(\sin^{-1}x)^2+A\cos^{-1}x+B$, where $A$ and $B$ are arbitrary constants, as its general solution.

Solution. Method: formation. Two constants, so the equation is of order two. Differentiate: $\dfrac{dy}{dx}=\dfrac{2\sin^{-1}x}{\sqrt{1-x^2}}-\dfrac{A}{\sqrt{1-x^2}}$, so $\sqrt{1-x^2}\,\dfrac{dy}{dx}=2\sin^{-1}x-A$. Differentiate again, applying the product rule on the left: $$\sqrt{1-x^2}\,\frac{d^2y}{dx^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}.$$ Multiplying by $\sqrt{1-x^2}$ removes the radicals, and $A$ and $B$ are both gone.

Question 14

Form the differential equation of all circles which pass through origin and whose centres lie on $y$-axis.

Solution. Method: formation. A circle with centre $(0,k)$ through the origin has radius $k$, so its equation is $x^2+y^2-2ky=0$, with one constant and hence a first-order equation. Differentiate: $2x+2y\dfrac{dy}{dx}-2k\dfrac{dy}{dx}=0$, so $k=\dfrac{x+y\,y’}{y’}$. Substitute into $x^2+y^2=2ky$: $$(x^2+y^2)\,y’=2y(x+y\,y’)=2xy+2y^2y’.$$ Collecting the $y’$ terms gives $(x^2-y^2)\dfrac{dy}{dx}-2xy=0$.

Question 15

Find the equation of a curve passing through origin and satisfying the differential equation $(1+x^2)\dfrac{dy}{dx}+2xy=4x^2$.

Solution. Method: linear, and again the left side is an exact derivative. Since $\dfrac{d}{dx}(1+x^2)=2x$, the left side is $\dfrac{d}{dx}\big[(1+x^2)y\big]$. So $\dfrac{d}{dx}\big[(1+x^2)y\big]=4x^2$, and integrating, $(1+x^2)y=\dfrac{4x^3}{3}+c$. The curve passes through the origin, so $c=0$. Thus $y=\dfrac{4x^3}{3(1+x^2)}$.

Question 16

Solve: $x^2\dfrac{dy}{dx}=x^2+xy+y^2$.

Solution. Method: homogeneous. Divide by $x^2$: $\dfrac{dy}{dx}=1+\dfrac yx+\left(\dfrac yx\right)^2$. Put $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. Then $v+x\dfrac{dv}{dx}=1+v+v^2$, which simplifies to $x\dfrac{dv}{dx}=1+v^2$. Separating, $\dfrac{dv}{1+v^2}=\dfrac{dx}{x}$, so $\tan^{-1}v=\log|x|+c$. Replacing $v$ by $\dfrac yx$ gives the result.

Question 17

Find the general solution of the differential equation $(1+y^2)+\left(x-e^{\tan^{-1}y}\right)\dfrac{dy}{dx}=0$.

Solution. Method: linear, with $x$ as the dependent variable. Divide by $\dfrac{dy}{dx}$ and by $1+y^2$ to turn the equation into $\dfrac{dx}{dy}+\dfrac{x}{1+y^2}=\dfrac{e^{\tan^{-1}y}}{1+y^2}$. Here $P=\dfrac{1}{1+y^2}$, so the integrating factor is $e^{\tan^{-1}y}$. Let $t=\tan^{-1}y$, so $dt=\dfrac{dy}{1+y^2}$. Then $$x\,e^{t}=\int e^{2t}\,dt=\frac{e^{2t}}{2}+c_1.$$ Multiplying by 2 and renaming the constant gives the general solution.

Question 18

Find the general solution of $y^2\,dx+(x^2-xy+y^2)\,dy=0$.

Solution. Method: homogeneous, with $x=vy$. Write $\dfrac{dx}{dy}=-\dfrac{x^2-xy+y^2}{y^2}$. Substituting $x=vy$, so $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$, we get $v+y\dfrac{dv}{dy}=-(v^2-v+1)$, hence $y\dfrac{dv}{dy}=-(v^2+1)$. Separating, $\dfrac{dv}{1+v^2}=-\dfrac{dy}{y}$, so $\tan^{-1}v=-\log y+c$. Therefore $\tan^{-1}\dfrac xy+\log y=c$. We take $x$ as the dependent variable here because the equation is simpler in that form: $y^2$ multiplies $dx$ alone.

Question 19

Solve: $(x+y)(dx-dy)=dx+dy$. [Hint: Substitute $x+y=z$ after separating $dx$ and $dy$.]

Solution. Method: substitution. Let $z=x+y$, so $dz=dx+dy$. The equation becomes $z\,(dx-dy)=dz$, that is $d(x-y)=\dfrac{dz}{z}$. Integrating, $x-y=\log|z|+C=\log|x+y|+C$. Equivalently, $x+y=ke^{x-y}$.

Question 20

Solve: $2(y+3)-xy\dfrac{dy}{dx}=0$, given that $y(1)=-2$.

Solution. Method: variables separable. Write $xy\,dy=2(y+3)\,dx$, so $\dfrac{y}{y+3}\,dy=\dfrac{2}{x}\,dx$. Since $\dfrac{y}{y+3}=1-\dfrac{3}{y+3}$, integrating gives $y-3\log|y+3|=2\log|x|+c$. At $x=1$, $y=-2$: $-2-3\log1=0+c$, so $c=-2$. Thus $$y-3\log|y+3|=2\log|x|-2,$$ which can be tidied to $x^2(y+3)^3=e^{y+2}$.

Question 21

Solve the differential equation $dy=\cos x\,(2-y\,\operatorname{cosec}x)\,dx$ given that $y=2$ when $x=\dfrac{\pi}{2}$.

Solution. Method: linear. Expand: $\dfrac{dy}{dx}=2\cos x-y\cos x\operatorname{cosec}x=2\cos x-y\cot x$. So $\dfrac{dy}{dx}+y\cot x=2\cos x$. The integrating factor is $e^{\int\cot x\,dx}=\sin x$. Then $\dfrac{d}{dx}(y\sin x)=2\sin x\cos x=\sin2x$, and integrating, $y\sin x=\sin^2x+c$. At $x=\dfrac\pi2$, $y=2$: $2=1+c$, so $c=1$. Hence $y\sin x=\sin^2x+1$, i.e. $y=\sin x+\operatorname{cosec}x$. The key prints $y\sin x=-\dfrac{\cos2x}{2}+\dfrac32$, which is the same equation, since $\sin^2x=\dfrac{1-\cos2x}{2}$.

Question 22

Form the differential equation by eliminating $A$ and $B$ in $Ax^2+By^2=1$.

Solution. Method: formation. Two constants, so order two. Differentiate: $2Ax+2By\,y’=0$, i.e. $Ax+By\,y’=0$. Differentiate again: $A+B\big[(y’)^2+y\,y”\big]=0$. From the first, $A=-\dfrac{By\,y’}{x}$. Substituting into the second and cancelling $B$ (which is non-zero): $$-\frac{y\,y’}{x}+(y’)^2+y\,y”=0.$$ Multiply by $x$ to clear the fraction.

Question 23

Solve the differential equation $(1+y^2)\tan^{-1}x\,dx+2y\,(1+x^2)\,dy=0$.

Solution. Method: variables separable. Divide by $(1+x^2)(1+y^2)$: $\dfrac{\tan^{-1}x}{1+x^2}\,dx+\dfrac{2y}{1+y^2}\,dy=0$. The first term is $\tan^{-1}x$ times its own derivative, so it integrates to $\dfrac12(\tan^{-1}x)^2$; the second is $\dfrac{d}{dy}\log(1+y^2)$. Therefore $\dfrac12(\tan^{-1}x)^2+\log(1+y^2)=c$.

Question 24

Find the differential equation of system of concentric circles with centre $(1,2)$.

Solution. Method: formation. Concentric circles about $(1,2)$ differ only in the radius $r$, so they form a one-parameter family $(x-1)^2+(y-2)^2=r^2$, and the equation is of order one. Differentiating, $2(x-1)+2(y-2)\dfrac{dy}{dx}=0$. The radius has dropped out.

Common mistakes

• Dropping the constant. A general solution needs $+C$, added at the moment of integration. Adding it after rearranging (for example after taking logarithms or exponentials) changes what the constant means, so write it immediately.
• Forgetting the modulus or the case split. In Questions 4, 8 and 20, $\int\dfrac{dx}{x}=\log|x|$, not $\log x$. In Question 6, the formula $\dfrac{e^{mx}}{m+a}$ fails at $m=-a$, where the solution has the form $(x+C)e^{-ax}$.
• Using the wrong dependent variable. Questions 10, 17 and 18 are awkward in $y$ but easy in $x$. If $\dfrac{dy}{dx}$ makes the equation non-linear, try $\dfrac{dx}{dy}$ before giving up.
• Miscounting constants in formation. The order of the equation equals the number of independent arbitrary constants. Question 14 has one constant, so a second derivative should not appear; Questions 13 and 22 have two, so it must.
• Sign slips in the integrating factor. In Question 12, $-\dfrac{t}{1+t}$ integrates to $-t+\log(1+t)$, giving $(1+t)e^{-t}$. Writing $+t$ instead gives the factor for a different equation.
• Applying the initial condition before the constant is isolated. In Questions 11, 20 and 21, substitute the given point into the general solution, solve for the constant, and only then evaluate the final value asked for.

Practise next

• The other bands of this unit: Differential Equations, long answer (Q25-33), objective, part 1 (Q34-54), objective, part 2 (Q55-75) and fill in the blanks and true or false (Q76-77).
• Textbook practice: Exercise 9.2, general and particular solutions, Exercise 9.3, variables separable, Exercise 9.4, homogeneous equations, Exercise 9.5, linear equations and the Miscellaneous Exercise.