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Differential Equations

277 Questions
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2021 JEE Mains MCQ
JEE Main 2021 (Online) 27th August Evening Shift
A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, $-$3) from the line 3x + 4y = 5, is given by :
A.
$10{{{d^2}y} \over {d{x^2}}} = 11$
B.
$11{{{d^2}x} \over {d{y^2}}} = 10$
C.
$10{{{d^2}x} \over {d{y^2}}} = 11$
D.
$11{{{d^2}y} \over {d{x^2}}} = 10$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 27th August Evening Shift
If the solution curve of the differential equation (2x $-$ 10y3)dy + ydx = 0, passes through the points (0, 1) and (2, $\beta$), then $\beta$ is a root of the equation :
A.
y5 $-$ 2y $-$ 2 = 0
B.
2y5 $-$ 2y $-$ 1 = 0
C.
2y5 $-$ y2 $-$ 2 = 0
D.
y5 $-$ y2 $-$ 1 = 0
2021 JEE Mains MCQ
JEE Main 2021 (Online) 27th August Morning Shift
Let y = y(x) be the solution of the differential equation

${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$ such that y(0) = 7. Then y($\pi$) is equal to :
A.
$2{e^{{\pi ^2}}} + 5$
B.
${e^{{\pi ^2}}} + 5$
C.
$3{e^{{\pi ^2}}} + 5$
D.
$7{e^{{\pi ^2}}} + 5$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 27th August Morning Shift
Let us consider a curve, y = f(x) passing through the point ($-$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
A.
${x^2} + 2xf(x) - 12 = 0$
B.
${x^3} + xf(x) + 12 = 0$
C.
${x^3} - 3xf(x) - 4 = 0$
D.
${x^2} + 2xf(x) + 4 = 0$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 26th August Evening Shift
Let y(x) be the solution of the differential equation

2x2 dy + (ey $-$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
A.
0
B.
2
C.
loge 2
D.
loge (2e)
2021 JEE Mains MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Let y = y(x) be a solution curve of the differential equation $(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$, $x \in \left( {0,{\pi \over 2}} \right)$. If $\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$, then the value of $y\left( {{\pi \over 4}} \right)$ is :
A.
$ - {\pi \over 4}$
B.
${\pi \over 4} - 1$
C.
${\pi \over 4} + 1$
D.
${\pi \over 4}$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 27th July Evening Shift
Let y = y(x) be the solution of the differential

equation (x $-$ x3)dy = (y + yx2 $-$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
A.
4
B.
12
C.
8
D.
16
2021 JEE Mains MCQ
JEE Main 2021 (Online) 27th July Morning Shift
Let y = y(x) be solution of the differential equation

${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$, with y(0) = 0.

If $y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$, then the value of $\alpha$ is equal to :
A.
$ - {1 \over 4}$
B.
${1 \over 4}$
C.
$2$
D.
$ - {1 \over 2}$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 25th July Evening Shift
Let y = y(x) be the solution of the differential

equation xdy = (y + x3 cosx)dx with y($\pi$) = 0, then $y\left( {{\pi \over 2}} \right)$ is equal to :
A.
${{{\pi ^2}} \over 4} + {\pi \over 2}$
B.
${{{\pi ^2}} \over 2} + {\pi \over 4}$
C.
${{{\pi ^2}} \over 2} - {\pi \over 4}$
D.
${{{\pi ^4}} \over 4} - {\pi \over 2}$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 25th July Morning Shift
Let y = y(x) be the solution of the differential equation ${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$

then, the minimum value of $y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$ is equal to :
A.
$\left( {2 - \sqrt 3 } \right) - {\log _e}2$
B.
$\left( {2 + \sqrt 3 } \right) + {\log _e}2$
C.
$\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$
D.
$\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Let y = y(x) be the solution of the differential equation $\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$, with $y\left( {{\pi \over 4}} \right) = 0$. Then, the value of ${(y(0) + 1)^2}$ is equal to :
A.
e1/2
B.
e$-$1/2
C.
e$-$1
D.
e
2021 JEE Mains MCQ
JEE Main 2021 (Online) 20th July Evening Shift
Let y = y(x) satisfies the equation ${{dy} \over {dx}} - |A| = 0$, for all x > 0, where $A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$. If $y(\pi ) = \pi + 2$, then the value of $y\left( {{\pi \over 2}} \right)$ is :
A.
${\pi \over 2} + {4 \over \pi }$
B.
${\pi \over 2} - {1 \over \pi }$
C.
${{3\pi } \over 2} - {1 \over \pi }$
D.
${\pi \over 2} - {4 \over \pi }$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 20th July Morning Shift
Let y = y(x) be the solution of the differential equation $x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$, $ - 1 \le x \le 1$, $y\left( {{1 \over 2}} \right) = {\pi \over 6}$. Then the area of the region bounded by the curves x = 0, $x = {1 \over {\sqrt 2 }}$ and y = y(x) in the upper half plane is :
A.
${1 \over 8}(\pi - 1)$
B.
${1 \over {12}}(\pi - 3)$
C.
${1 \over 4}(\pi - 2)$
D.
${1 \over 6}(\pi - 1)$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 20th July Morning Shift
Let y = y(x) be the solution of the differential equation ${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$, y(1) = $-$1. Then the value of (y(3))2 is equal to :
A.
1 $-$ 4e3
B.
1 $-$ 4e6
C.
1 + 4e3
D.
1 + 4e6
2021 JEE Mains MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Let y = y(x) be the solution of the differential equation

${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$, 0 < x < 2.1, with y(2) = 0. Then the value of ${{dy} \over {dx}}$ at x = 1 is equal to :
A.
${{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}$
B.
${{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}$
C.
$ - {{2{e^2}} \over {{{(1 + {e^2})}^2}}}$
D.
${{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 18th March Morning Shift
The differential equation satisfied by the system of parabolas

y2 = 4a(x + a) is :
A.
$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0$
B.
$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0$
C.
$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$
D.
$y\left( {{{dy} \over {dx}}} \right) + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 17th March Evening Shift
If the curve y = y(x) is the solution of the differential equation

$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$, x > 0 which

passes through the point $\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$, then the value of y(16) is equal to :
A.
$4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$
B.
$\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$
C.
$\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$
D.
$4\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Let y = y(x) be the solution of the differential equation

$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$. Then, $y\left( {{\pi \over 3}} \right)$ is equal to :
A.
$2{\log _e}\left( {{{\sqrt 3 + 7} \over 2}} \right)$
B.
$2{\log _e}\left( {{{3\sqrt 3 - 8} \over 4}} \right)$
C.
$2{\log _e}\left( {{{2\sqrt 3 + 10} \over {11}}} \right)$
D.
$2{\log _e}\left( {{{2\sqrt 3 + 9} \over 6}} \right)$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 17th March Morning Shift
Which of the following is true for y(x) that satisfies the differential equation

${{dy} \over {dx}}$ = xy $-$ 1 + x $-$ y; y(0) = 0 :
A.
y(1) = 1
B.
y(1) = e$-$${1 \over 2}$ $-$ 1
C.
y(1) = e${1 \over 2}$ $-$ e$-$${1 \over 2}$
D.
y(1) = e${1 \over 2}$ $-$ 1
2021 JEE Mains MCQ
JEE Main 2021 (Online) 16th March Evening Shift
If y = y(x) is the solution of the differential equation

${{dy} \over {dx}}$ + (tan x) y = sin x, $0 \le x \le {\pi \over 3}$, with y(0) = 0, then $y\left( {{\pi \over 4}} \right)$ equal to :
A.
${1 \over 2}$loge 2
B.
$\left( {{1 \over {2\sqrt 2 }}} \right)$ loge 2
C.
loge 2
D.
${1 \over 4}$ loge 2
2021 JEE Mains MCQ
JEE Main 2021 (Online) 16th March Evening Shift
Let C1 be the curve obtained by the solution of differential equation

$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$. Let the curve C2 be the

solution of ${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :
A.
${\pi \over 4}$ + 1
B.
$\pi$ + 1
C.
$\pi$ $-$ 1
D.
${\pi \over 2}$ $-$ 1
2021 JEE Mains MCQ
JEE Main 2021 (Online) 16th March Morning Shift
If y = y(x) is the solution of the differential equation,

${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$, then the maximum value of the function y(x) over R is equal to:
A.
${1 \over 8}$
B.
8
C.
$-$${15 \over 4}$
D.
${1 \over 2}$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after ${k \over {{{\log }_e}\left( {{6 \over 5}} \right)}}$ hours, then ${\left( {{k \over {{{\log }_e}2}}} \right)^2}$ is equal to :
A.
16
B.
8
C.
2
D.
4
2021 JEE Mains MCQ
JEE Main 2021 (Online) 25th February Morning Shift
If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is ${{{x^2} - 4x + y + 8} \over {x - 2}}$, then this curve also passes through the point :
A.
(4, 4)
B.
(5, 5)
C.
(5, 4)
D.
(4, 5)
2021 JEE Mains MCQ
JEE Main 2021 (Online) 24th February Evening Shift
If a curve y = f(x) passes through the point (1, 2) and satisfies $x {{dy} \over {dx}} + y = b{x^4}$, then for what value of b, $\int\limits_1^2 {f(x)dx = {{62} \over 5}} $?
A.
${{31} \over 5}$
B.
10
C.
5
D.
${{62} \over 5}$
2021 JEE Mains MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The population P = P(t) at time 't' of a certain species follows the differential equation

${{dP} \over {dt}}$ = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :
A.
${\log _e}18$
B.
${1 \over 2}{\log _e}18$
C.
2${\log _e}18$
D.
${\log _e}9$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 6th September Evening Slot
If $y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$ is the solution of the differential equation,

${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$,

$0 < x < {\pi \over 2}$, then the function p(x) is equal to :
A.
cot x
B.
sec x
C.
tan x
D.
cosec x
2020 JEE Mains MCQ
JEE Main 2020 (Online) 6th September Morning Slot
The general solution of the differential equation

$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $ + xy${{dy} \over {dx}}$ = 0 is :

(where C is a constant of integration)
A.
$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$
B.
$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$
C.
$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$
D.
$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Let y = y(x) be the solution of the differential equation

cosx${{dy} \over {dx}}$ + 2ysinx = sin2x, x $ \in $ $\left( {0,{\pi \over 2}} \right)$.

If y$\left( {{\pi \over 3}} \right)$ = 0, then y$\left( {{\pi \over 4}} \right)$ is equal to :
A.
${1 \over {\sqrt 2 }} - 1$
B.
${\sqrt 2 - 2}$
C.
${2 - \sqrt 2 }$
D.
${2 + \sqrt 2 }$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 5th September Morning Slot
If y = y(x) is the solution of the differential

equation ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$ satisfying y(0) = 1, then a value of y(loge13) is :
A.
-1
B.
1
C.
0
D.
2
2020 JEE Mains MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The solution of the differential equation

${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$ is:

(where c is a constant of integration)
A.
$x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$
B.
$y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$
C.
x – loge(y+3x) = C
D.
x – 2loge(y+3x) = C
2020 JEE Mains MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Let y = y(x) be the solution of the differential equation,
xy'- y = x2(xcosx + sinx), x > 0. if y ($\pi $) = $\pi $ then
$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$ is equal to :
A.
$1 + {\pi \over 2}$
B.
$2 + {\pi \over 2} + {{{\pi ^2}} \over 4}$
C.
$2 + {\pi \over 2}$
D.
$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
If x3dy + xy dx = x2dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :
A.
${{\sqrt e } \over 2}$
B.
${1 \over 2} + \sqrt e $
C.
${3 \over 2} + \sqrt e $
D.
${3 \over 2}\sqrt e $
2020 JEE Mains MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The solution curve of the differential equation,

(1 + e-x)(1 + y2)${{dy} \over {dx}}$ = y2,

which passes through the point (0, 1), is :
A.
y2 + 1 = y$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$
B.
y2 + 1 = y$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$
C.
y2 = 1 + ${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$
D.
y2 = 1 + ${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x2dy= (2xy + y2)dx, then $f\left( {{1 \over 2}} \right)$ is equal to :
A.
${1 \over {1 - {{\log }_e}2}}$
B.
${1 \over {1 + {{\log }_e}2}}$
C.
${{ - 1} \over {1 + {{\log }_e}2}}$
D.
${1 + {{\log }_e}2}$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
Let y = y(x) be the solution of the differential equation,
${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$, y > 0,y(0) = 1.
If y($\pi $) = a and ${{dy} \over {dx}}$ at x = $\pi $ is b, then the ordered pair (a, b) is equal to :
A.
(2, 1)
B.
$\left( {2,{3 \over 2}} \right)$
C.
(1, -1)
D.
(1, 1)
2020 JEE Mains MCQ
JEE Main 2020 (Online) 9th January Evening Slot
If ${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$; y(1) = 1; then a value of x satisfying y(x) = e is :
A.
$\sqrt 2 e$
B.
${1 \over 2}\sqrt 3 e$
C.
${e \over {\sqrt 2 }}$
D.
$\sqrt 3 e$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 8th January Evening Slot
The differential equation of the family of curves, x2 = 4b(y + b), b $ \in $ R, is :
A.
x(y')2 = x – 2yy'
B.
x(y')2 = 2yy' – x
C.
xy" = y'
D.
x(y')2 = x + 2yy'
2020 JEE Mains MCQ
JEE Main 2020 (Online) 8th January Morning Slot
Let y = y(x) be a solution of the differential equation,

$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$, |x| < 1.

If $y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$, then $y\left( { - {1 \over {\sqrt 2 }}} \right)$ is equal to :
A.
$ - {{\sqrt 3 } \over 2}$
B.
None of those
C.
${{1 \over {\sqrt 2 }}}$
D.
$-{{1 \over {\sqrt 2 }}}$
2020 JEE Mains MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Let y = y(x) be the solution curve of the differential equation,

$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
A.
2 + e
B.
-e
C.
2
D.
2 - e
2020 JEE Mains MCQ
JEE Main 2020 (Online) 7th January Morning Slot
If y = y(x) is the solution of the differential equation, ${e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}$ such that y(0) = 0, then y(1) is equal to:
A.
2 + loge2
B.
loge2
C.
1 + loge2
D.
2e
2019 JEE Mains MCQ
JEE Main 2019 (Online) 12th April Evening Slot
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 JEE Mains MCQ
JEE Main 2019 (Online) 12th April Morning Slot
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 JEE Mains MCQ
JEE Main 2019 (Online) 10th April Evening Slot
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 :
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 JEE Mains MCQ
JEE Main 2019 (Online) 10th April Morning Slot
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 :
A.
${1 \over 2} - e$
B.
$e - 2$
C.
$2 + {1 \over e}$
D.
${1 \over e} - 2$
2019 JEE Mains MCQ
JEE Main 2019 (Online) 9th April Evening Slot
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 :-
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 JEE Mains MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The solution of the differential equation

$x{{dy} \over {dx}} + 2y$ = 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 JEE Mains MCQ
JEE Main 2019 (Online) 8th April Morning Slot
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 :
A.
${1 \over 2}$
B.
${1 \over 16}$
C.
1
D.
${1 \over 4}$
2019 JEE Mains MCQ
JEE Main 2019 (Online) 12th January Evening Slot
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 JEE Mains MCQ
JEE Main 2019 (Online) 12th January Morning Slot
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.
${e \over 4}$