Differential Equations
246 Questions
Start JEE Mains Test
2020
Q201
JEE Mains
Numerical
14 Mar 2026
If for x $ \ge $ 0, y = y(x) is the solution of the
differential equation
(x + 1)dy = ((x + 1)2 + y – 3)dx, y(2) = 0, then y(3) is equal to _______.
(x + 1)dy = ((x + 1)2 + y – 3)dx, y(2) = 0, then y(3) is equal to _______.
Correct Answer: 3
Explanation:
(x + 1)dy = ((x + 1)2 + y – 3)dx
$ \Rightarrow $ (1 + x)${{dy} \over {dx}}$ - y = (1 + x)2 - 3
$ \Rightarrow $ ${{dy} \over {dx}} - {y \over {1 + x}} = \left( {1 + x} \right) - {3 \over {1 + x}}$
I.F = ${e^{ - \int {{{dx} \over {1 + x}}} }}$ = ${1 \over {1 + x}}$
Solution of the differential equation,
$y\left( {{1 \over {1 + x}}} \right)$ = $\int {\left( {\left( {1 + x} \right) - {3 \over {1 + x}}} \right)\left( {{1 \over {1 + x}}} \right)dx} $
$ \Rightarrow $ ${y \over {1 + x}}$ = $\int {{{{x^2} + 2x + 1 - 3} \over {{{\left( {x + 1} \right)}^2}}}dx} $
$ \Rightarrow $ ${y \over {1 + x}}$ = x + ${3 \over {1 + x}}$ + C
As y(2) = 0 $ \Rightarrow $ x = 2, y = 0
$ \therefore $ 0 = 2 + ${3 \over {1 + 2}}$ + C
$ \Rightarrow $ C = -3
So solution is ${y \over {1 + x}}$ = x + ${3 \over {1 + x}}$ - 3
y(3) means x = 3 and find value of y.
${y \over {1 + 3}} = 3 + {3 \over {1 + 3}} - 3$
$ \Rightarrow $ y = 3
$ \Rightarrow $ (1 + x)${{dy} \over {dx}}$ - y = (1 + x)2 - 3
$ \Rightarrow $ ${{dy} \over {dx}} - {y \over {1 + x}} = \left( {1 + x} \right) - {3 \over {1 + x}}$
I.F = ${e^{ - \int {{{dx} \over {1 + x}}} }}$ = ${1 \over {1 + x}}$
Solution of the differential equation,
$y\left( {{1 \over {1 + x}}} \right)$ = $\int {\left( {\left( {1 + x} \right) - {3 \over {1 + x}}} \right)\left( {{1 \over {1 + x}}} \right)dx} $
$ \Rightarrow $ ${y \over {1 + x}}$ = $\int {{{{x^2} + 2x + 1 - 3} \over {{{\left( {x + 1} \right)}^2}}}dx} $
$ \Rightarrow $ ${y \over {1 + x}}$ = x + ${3 \over {1 + x}}$ + C
As y(2) = 0 $ \Rightarrow $ x = 2, y = 0
$ \therefore $ 0 = 2 + ${3 \over {1 + 2}}$ + C
$ \Rightarrow $ C = -3
So solution is ${y \over {1 + x}}$ = x + ${3 \over {1 + x}}$ - 3
y(3) means x = 3 and find value of y.
${y \over {1 + 3}} = 3 + {3 \over {1 + 3}} - 3$
$ \Rightarrow $ y = 3
2019
Q202
JEE Mains
MCQ
14 Mar 2026
The general solution of the differential equation (y2
– x3)dx – xydy = 0 (x $ \ne $ 0) is :
(where c is a constant of integration)
A.
y2
+ 2x3
+ cx2
= 0
B.
y2
+ 2x2
+ cx3
= 0
C.
y2
– 2x + cx3
= 0
D.
y2
– 2x3
+ cx2
= 0
2019
Q203
JEE Mains
MCQ
14 Mar 2026
Consider the differential equation, ${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$, If value of y is 1 when x = 1, then the value of x
for which y = 2, is :
A.
${3 \over 2} - {1 \over {\sqrt e }}$
B.
${1 \over 2} + {1 \over {\sqrt e }}$
C.
${5 \over 2} + {1 \over {\sqrt e }}$
D.
${3 \over 2} - \sqrt e $
2019
Q204
JEE Mains
MCQ
14 Mar 2026
Let y = y(x) be the solution of the differential equation,
${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$, such that y(0) = 1. Then :
${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$, such that y(0) = 1. Then :
A.
$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2 $
B.
$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $
C.
$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$
D.
$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2 $
2019
Q205
JEE Mains
MCQ
14 Mar 2026
If y = y(x) is the solution of the differential equation
${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$,
such that y (0) = 0, then $y\left( { - {\pi \over 4}} \right)$ is equal to :
${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$,
such that y (0) = 0, then $y\left( { - {\pi \over 4}} \right)$ is equal to :
A.
${1 \over 2} - e$
B.
$e - 2$
C.
$2 + {1 \over e}$
D.
${1 \over e} - 2$
2019
Q206
JEE Mains
MCQ
14 Mar 2026
If $\cos x{{dy} \over {dx}} - y\sin x = 6x$, (0 < x < ${\pi \over 2}$)
and $y\left( {{\pi \over 3}} \right)$ = 0 then $y\left( {{\pi \over 6}} \right)$ is equal to :-
and $y\left( {{\pi \over 3}} \right)$ = 0 then $y\left( {{\pi \over 6}} \right)$ is equal to :-
A.
$ - {{{\pi ^2}} \over {2 }}$
B.
$ - {{{\pi ^2}} \over {4\sqrt 3 }}$
C.
$ {{{\pi ^2}} \over {2\sqrt 3 }}$
D.
$ - {{{\pi ^2}} \over {2\sqrt 3 }}$
2019
Q207
JEE Mains
MCQ
14 Mar 2026
The solution of the differential equation
$x{{dy} \over {dx}} + 2y$ = x2 (x $ \ne $ 0) with y(1) = 1, is :
$x{{dy} \over {dx}} + 2y$ = x2 (x $ \ne $ 0) with y(1) = 1, is :
A.
$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$
B.
$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$
C.
$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$
D.
$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$
2019
Q208
JEE Mains
MCQ
14 Mar 2026
Let y = y(x) be the solution of the differential equation,
${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$
such that y(0) = 0. If $\sqrt ay(1)$ = $\pi \over 32$ , then the value of 'a' is :
${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$
such that y(0) = 0. If $\sqrt ay(1)$ = $\pi \over 32$ , then the value of 'a' is :
A.
${1 \over 2}$
B.
${1 \over 16}$
C.
1
D.
${1 \over 4}$
2019
Q209
JEE Mains
MCQ
14 Mar 2026
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as ${{{x^2} - 2y} \over x}$, then the curve also passes through the point :
A.
(–1, 2)
B.
$\left( { - \sqrt 2 ,1} \right)$
C.
$\left( { \sqrt 3 ,0} \right)$
D.
(3, 0)
2019
Q210
JEE Mains
MCQ
14 Mar 2026
Let y = y(x) be the solution of the differential equation, x${{dy} \over {dx}}$ + y = x loge x, (x > 1). If 2y(2) = loge 4 $-$ 1, then y(e) is equal to :
A.
$ - {e \over 2}$
B.
$ - {{{e^2}} \over 2}$
C.
${{{e^2}} \over 4}$
D.
${e \over 4}$
2019
Q211
JEE Mains
MCQ
14 Mar 2026
The solution of the differential equation,
${{dy} \over {dx}}$ = (x – y)2, when y(1) = 1, is :
${{dy} \over {dx}}$ = (x – y)2, when y(1) = 1, is :
A.
$-$ loge $\left| {{{1 + x - y} \over {1 - x + y}}} \right|$ = x + y $-$ 2
B.
loge $\left| {{{2 - x} \over {2 - y}}} \right|$ = x $-$ y
C.
loge $\left| {{{2 - y} \over {2 - x}}} \right|$ = 2(y $-$ 1)
D.
$-$ loge $\left| {{{1 - x + y} \over {1 + x - y}}} \right|$ = 2(x $-$ 1)
2019
Q212
JEE Mains
MCQ
14 Mar 2026
If y(x) is the solution of the differential equation ${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$ where $y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$ then
A.
y(loge2) = loge4
B.
y(x) is decreasing in (0, 1)
C.
y(loge2) = ${{{{\log }_e}2} \over 4}$
D.
y(x) is decreasing in $\left( {{1 \over 2},1} \right)$
2019
Q213
JEE Mains
MCQ
14 Mar 2026
Let f be a differentiable function such that f '(x) = 7 - ${3 \over 4}{{f\left( x \right)} \over x},$ (x > 0) and f(1) $ \ne $ 4. Then $\mathop {\lim }\limits_{x \to 0'} \,$ xf$\left( {{1 \over x}} \right)$ :
A.
does not exist
B.
exists and equals ${4 \over 7}$
C.
exists and equals 4
D.
exists and equals 0
2019
Q214
JEE Mains
MCQ
14 Mar 2026
The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is :
A.
a circle with centre on the y-axis
B.
an ellipse with major axis along the y-axis
C.
a circle with centre on the x-axis
D.
a hyperbola with transverse axis along the x-axis
2019
Q215
JEE Mains
MCQ
14 Mar 2026
If ${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$ and $y\left( {{\pi \over 4}} \right) = {4 \over 3},$ then $y\left( { - {\pi \over 4}} \right)$ equals -
A.
${1 \over 3} + {e^6}$
B.
${1 \over 3}$
C.
${1 \over 3}$ + e3
D.
$-$ ${4 \over 3}$
2019
Q216
JEE Mains
MCQ
14 Mar 2026
Let f : [0,1] $ \to $ R be such that f(xy) = f(x).f(y), for all x, y $ \in $ [0, 1], and f(0) $ \ne $ 0. If y = y(x) satiesfies the differential equation, ${{dy} \over {dx}}$ = f(x) with y(0) = 1, then y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ is equal to :
A.
3
B.
4
C.
2
D.
5
2019
Q217
JEE Mains
MCQ
14 Mar 2026
If y = y(x) is the solution of the differential equation,
x$dy \over dx$ + 2y = x2, satisfying y(1) = 1, then y($1\over2$) is equal to :
x$dy \over dx$ + 2y = x2, satisfying y(1) = 1, then y($1\over2$) is equal to :
A.
$ {{7} \over {64}}$
B.
$ {{49} \over {16}}$
C.
$ {{1} \over {4}}$
D.
$ {{13} \over {16}}$
2018
Q218
JEE Mains
MCQ
14 Mar 2026
The differential equation representing the family of ellipse having foci eith on the x-axis or on the $y$-axis, center at the origin and passing through the point (0, 3) is :
A.
xy y'' + x (y')2 $-$ y y' = 0
B.
x + y y'' = 0
C.
xy y'+ y2 $-$ 9 = 0
D.
xy y' $-$ y2 + 9 = 0
2018
Q219
JEE Mains
MCQ
14 Mar 2026
Let y = y(x) be the solution of the differential equation
$\sin x{{dy} \over {dx}} + y\cos x = 4x$, $x \in \left( {0,\pi } \right)$.
If $y\left( {{\pi \over 2}} \right) = 0$, then $y\left( {{\pi \over 6}} \right)$ is equal to :
$\sin x{{dy} \over {dx}} + y\cos x = 4x$, $x \in \left( {0,\pi } \right)$.
If $y\left( {{\pi \over 2}} \right) = 0$, then $y\left( {{\pi \over 6}} \right)$ is equal to :
A.
$ - {4 \over 9}{\pi ^2}$
B.
${4 \over {9\sqrt 3 }}{\pi ^2}$
C.
$ - {8 \over {9\sqrt 3 }}{\pi ^2}$
D.
$ - {8 \over 9}{\pi ^2}$
2018
Q220
JEE Mains
MCQ
14 Mar 2026
The curve satifying the differeial equation, (x2 $-$ y2) dx + 2xydy = 0 and passing through the point (1, 1) is :
A.
a circle of radius one.
B.
a hyperbola.
C.
an ellipse.
D.
a circle of radius two.
2018
Q221
JEE Mains
MCQ
14 Mar 2026
Let y = y(x) be the solution of the differential equation ${{dy} \over {dx}} + 2y = f\left( x \right),$
where $f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$
If y(0) = 0, then $y\left( {{3 \over 2}} \right)$ is :
where $f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$
If y(0) = 0, then $y\left( {{3 \over 2}} \right)$ is :
A.
${{{e^2} + 1} \over {2{e^4}}}$
B.
${1 \over {2e}}$
C.
${{{e^2} - 1} \over {{e^3}}}$
D.
${{{e^2} - 1} \over {2{e^3}}}$
2017
Q222
JEE Mains
MCQ
14 Mar 2026
If 2x = y${^{{1 \over 5}}}$ + y${^{ - {1 \over 5}}}$ and
(x2 $-$ 1) ${{{d^2}y} \over {d{x^2}}}$ + $\lambda $x ${{dy} \over {dx}}$ + ky = 0,
then $\lambda $ + k is equal to :
(x2 $-$ 1) ${{{d^2}y} \over {d{x^2}}}$ + $\lambda $x ${{dy} \over {dx}}$ + ky = 0,
then $\lambda $ + k is equal to :
A.
$-$ 23
B.
$-$ 24
C.
26
D.
$-$ 26
2017
Q223
JEE Mains
MCQ
14 Mar 2026
The curve satisfying the differential equation, ydx $-$(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
A.
$\left( {{1 \over 4}, - {1 \over 2}} \right)$
B.
$\left( { - {1 \over 3},{1 \over 3}} \right)$
C.
$\left( {{1 \over 3}, - {1 \over 3}} \right)$
D.
$\left( {{1 \over 4}, {1 \over 2}} \right)$
2017
Q224
JEE Mains
MCQ
14 Mar 2026
If $\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$ and y(0) = 1,
then $y\left( {{\pi \over 2}} \right)$ is equal to :
then $y\left( {{\pi \over 2}} \right)$ is equal to :
A.
$ - {2 \over 3}$
B.
$ - {1 \over 3}$
C.
${4 \over 3}$
D.
${1 \over 3}$
2016
Q225
JEE Mains
MCQ
14 Mar 2026
The solution of the differential equation
${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$
where 0 $ \le $ x < ${\pi \over 2}$, and y (0) = 1, is given by :
${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$
where 0 $ \le $ x < ${\pi \over 2}$, and y (0) = 1, is given by :
A.
y = 1 $-$ ${x \over {\sec x + \tan x}}$
B.
y2 = 1 + ${x \over {\sec x + \tan x}}$
C.
y2 = 1 $-$ ${x \over {\sec x + \tan x}}$
D.
y = 1 + ${x \over {\sec x + \tan x}}$
2016
Q226
JEE Mains
MCQ
14 Mar 2026
If f(x) is a differentiable function in the interval (0, $\infty $) such that f (1) = 1 and
$\mathop {\lim }\limits_{t \to x} $ ${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$ for each x > 0, then $f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$ equal to :
$\mathop {\lim }\limits_{t \to x} $ ${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$ for each x > 0, then $f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$ equal to :
A.
${{13} \over 6}$
B.
${{23} \over 18}$
C.
${{25} \over 9}$
D.
${{31} \over 18}$
2016
Q227
JEE Mains
MCQ
14 Mar 2026
If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y(1+xy) dx=x$ $dy$, then $f\left( { - {1 \over 2}} \right)$ is equal to :
A.
${2 \over 5}$
B.
${4 \over 5}$
C.
$-{2 \over 5}$
D.
$-{4 \over 5}$
2015
Q228
JEE Mains
MCQ
14 Mar 2026
Let $y(x)$ be the solution of the differential equation
$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$ Then $y(e)$ is equal to :
$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$ Then $y(e)$ is equal to :
A.
$2$
B.
$2e$
C.
$e$
D.
$0$
2014
Q229
JEE Mains
MCQ
14 Mar 2026
Let the population of rabbits surviving at time $t$ be governed by the differential equation ${{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200.$ If $p(0)=100,$ then $p(t)$ equals:
A.
$600 - 500\,{e^{t/2}}$
B.
$400 - 300\,{e^{-t/2}}$
C.
$400 - 300\,{e^{t/2}}$
D.
$300 - 200\,{e^{-t/2}}$
2013
Q230
JEE Mains
MCQ
14 Mar 2026
At present, a firm is manufacturing $2000$ items. It is estimated that the rate of change of production P w.r.t. additional number of workers $x$ is given by ${{dp} \over {dx}} = 100 - 12\sqrt x .$ If the firm employs $25$ more workers, then the new level of production of items is
A.
$2500$
B.
$3000$
C.
$3500$
D.
$4500$
2012
Q231
JEE Mains
MCQ
14 Mar 2026
The population $p$ $(t)$ at time $t$ of a certain mouse species satisfies the differential equation ${{dp\left( t \right)} \over {dt}} = 0.5\,p\left( t \right) - 450.\,\,$ If $p(0)=850,$ then the time at which the population becomes zero is :
A.
$2ln$ $18$
B.
$ln$ $9$
C.
${1 \over 2}$$ln$ $18$
D.
$ln$ $18$
2011
Q232
JEE Mains
MCQ
14 Mar 2026
Let $I$ be the purchase value of an equipment and $V(t)$ be the value after it has been used for $t$ years. The value $V(t)$ depreciates at a rate given by differential equation ${{dv\left( t \right)} \over {dt}} = - k\left( {T - t} \right),$ where $k>0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V(T)$ of the equipment is
A.
$I - {{k{T^2}} \over 2}$
B.
$I - {{k{{\left( {T - t} \right)}^2}} \over 2}$
C.
${e^{ - kT}}$
D.
${T^2} - {1 \over k}$
2011
Q233
JEE Mains
MCQ
14 Mar 2026
If ${{dy} \over {dx}} = y + 3 > 0\,\,$ and $y(0)=2,$ then $y\left( {\ln 2} \right)$ is equal to :
A.
$5$
B.
$13$
C.
$-2$
D.
$7$
2010
Q234
JEE Mains
MCQ
14 Mar 2026
Solution of the differential equation
$\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi \over 2}$ is :
$\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi \over 2}$ is :
A.
$y\sec x = \tan x + c$
B.
$y\tan x = \sec x + c$
C.
$\tan x = \left( {\sec x + c} \right)y$
D.
$\sec x = \left( {\tan x + c} \right)y$
2009
Q235
JEE Mains
MCQ
14 Mar 2026
The differential equation which represents the family of curves $y = {c_1}{e^{{c_2}x}},$ where ${c_1}$ , and ${c_2}$ are arbitrary constants, is
A.
$y'' = y'y$
B.
$yy'' = y'$
C.
$yy'' = {\left( {y'} \right)^2}$
D.
$y' = {y^2}$
2008
Q236
JEE Mains
MCQ
14 Mar 2026
The solution of the differential equation
${{dy} \over {dx}} = {{x + y} \over x}$ satisfying the condition $y(1)=1$ is :
${{dy} \over {dx}} = {{x + y} \over x}$ satisfying the condition $y(1)=1$ is :
A.
$y = \ln x + x$
B.
$y = x\ln x + {x^2}$
C.
$y = x{e^{\left( {x - 1} \right)}}\,$
D.
$y = x\,\ln x + x$
2007
Q237
JEE Mains
MCQ
14 Mar 2026
The differential equation of all circles passing through the origin and having their centres on the $x$-axis is :
A.
${y^2} = {x^2} + 2xy{{dy} \over {dx}}$
B.
${y^2} = {x^2} - 2xy{{dy} \over {dx}}$
C.
${x^2} = {y^2} + xy{{dy} \over {dx}}$
D.
${x^2} = {y^2} + 3xy{{dy} \over {dx}}$
2006
Q238
JEE Mains
MCQ
14 Mar 2026
The differential equation whose solution is $A{x^2} + B{y^2} = 1$
where $A$ and $B$ are arbitrary constants is of
where $A$ and $B$ are arbitrary constants is of
A.
second order and second degree
B.
first order and second degree
C.
first order and first degree
D.
second order and first degree
2005
Q239
JEE Mains
MCQ
14 Mar 2026
The differential equation representing the family of curves ${y^2} = 2c\left( {x + \sqrt c } \right),$ where $c>0,$ is a parameter, is of order and degree as follows:
A.
order $1,$ degree $2$
B.
order $1,$ degree $1$
C.
order $1,$ degree $3$
D.
order $2,$ degree $2$
2005
Q240
JEE Mains
MCQ
14 Mar 2026
If $x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),$ then the solution of the equation is :
A.
$y\log \left( {{x \over y}} \right) = cx$
B.
$x\log \left( {{y \over x}} \right) = cy$
C.
$\log \left( {{y \over x}} \right) = cx$
D.
$\log \left( {{x \over y}} \right) = cy$
2004
Q241
JEE Mains
MCQ
14 Mar 2026
The differential equation for the family of circle ${x^2} + {y^2} - 2ay = 0,$ where a is an arbitrary constant is :
A.
$\left( {{x^2} + {y^2}} \right)y' = 2xy$
B.
$2\left( {{x^2} + {y^2}} \right)y' = xy$
C.
$\left( {{x^2} - {y^2}} \right)y' =2 xy$
D.
$2\left( {{x^2} - {y^2}} \right)y' = xy$
2004
Q242
JEE Mains
MCQ
14 Mar 2026
Solution of the differential equation $ydx + \left( {x + {x^2}y} \right)dy = 0$ is
A.
$log$ $y=Cx$
B.
$ - {1 \over {xy}} + \log y = C$
C.
${1 \over {xy}} + \log y = C$
D.
$ - {1 \over {xy}} = C$
2003
Q243
JEE Mains
MCQ
14 Mar 2026
The degree and order of the differential equation of the family of all parabolas whose axis is $x$-axis, are respectively.
A.
$2, 3$
B.
$2,1$
C.
$1,2$
D.
$3,2.$
2003
Q244
JEE Mains
MCQ
14 Mar 2026
The solution of the differential equation
$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$ is :
$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$ is :
A.
$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$
B.
$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$
C.
$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$
D.
$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$
2002
Q245
JEE Mains
MCQ
14 Mar 2026
The order and degree of the differential equation
$\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}$ are
$\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}$ are
A.
$\left( {1,{2 \over 3}} \right)$
B.
$(3, 1)$
C.
$(3,3)$
D.
$(1,2)$
2002
Q246
JEE Mains
MCQ
14 Mar 2026
The solution of the equation $\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}$
A.
${{{e^{ - 2x}}} \over 4}$
B.
${{{e^{ - 2x}}} \over 4} + cx + d$
C.
${1 \over 4}{e^{ - 2x}} + c{x^2} + d$
D.
$\,{1 \over 4}{e^{ - 4x}} + cx + d$