Differential Equations
234 Questions
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 :
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 :
$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$, 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 :
$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$. 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 :
${{dy} \over {dx}}$ = 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 :
${{dy} \over {dx}}$ + (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 :
$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$. 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:
${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$, 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 :
${{dP} \over {dt}}$ = 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.
${\log _e}9$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
If ${y^{1/4}} + {y^{ - 1/4}} = 2x$, and
$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$, then | $\alpha$ $-$ $\beta$ | is equal to __________.
$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$, then | $\alpha$ $-$ $\beta$ | is equal to __________.
Correct Answer: 17
Explanation:
${y^{{1 \over 4}}} + {1 \over {{y^{{1 \over 4}}}}} = 2x$
$ \Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$
$ \Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1} $ or $x - \sqrt {{x^2} - 1} $
So, ${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = 1 + {x \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow {1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = {{{y^{{1 \over 4}}}} \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow {{dy} \over {dx}} = {{4y} \over {\sqrt {{x^2} - 1} }}$ .... (1)
Hence, ${{{d^2}y} \over {d{x^2}}} = 4{{\left( {\sqrt {{x^2} - 1} } \right)y' - {{yx} \over {\sqrt {{x^2} - 1} }}} \over {{x^2} - 1}}$
$ \Rightarrow ({x^2} - 1)y'' = 4{{({x^2} - 1)y' - xy} \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow ({x^2} - 1)y'' = 4\left( {\sqrt {{x^2} - 1} y' - {{xy} \over {\sqrt {{x^2} - 1} }}} \right)$
$ \Rightarrow ({x^2} - 1)y'' = 4\left( {4y - {{xy'} \over 4}} \right)$ (from I)
$ \Rightarrow ({x^2} - 1)y'' + xy' - 16y = 0$
So, | $\alpha$ $-$ $\beta$ | = 17
$ \Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$
$ \Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1} $ or $x - \sqrt {{x^2} - 1} $
So, ${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = 1 + {x \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow {1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = {{{y^{{1 \over 4}}}} \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow {{dy} \over {dx}} = {{4y} \over {\sqrt {{x^2} - 1} }}$ .... (1)
Hence, ${{{d^2}y} \over {d{x^2}}} = 4{{\left( {\sqrt {{x^2} - 1} } \right)y' - {{yx} \over {\sqrt {{x^2} - 1} }}} \over {{x^2} - 1}}$
$ \Rightarrow ({x^2} - 1)y'' = 4{{({x^2} - 1)y' - xy} \over {\sqrt {{x^2} - 1} }}$
$ \Rightarrow ({x^2} - 1)y'' = 4\left( {\sqrt {{x^2} - 1} y' - {{xy} \over {\sqrt {{x^2} - 1} }}} \right)$
$ \Rightarrow ({x^2} - 1)y'' = 4\left( {4y - {{xy'} \over 4}} \right)$ (from I)
$ \Rightarrow ({x^2} - 1)y'' + xy' - 16y = 0$
So, | $\alpha$ $-$ $\beta$ | = 17
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
Let y = y(x) be the solution of the differential equation dy = e$\alpha$x + y dx; $\alpha$ $\in$ N. If y(loge2) = loge2 and y(0) = loge$\left( {{1 \over 2}} \right)$, then the value of $\alpha$ is equal to _____________.
Correct Answer: 2
Explanation:
$\int {{e^{ - y}}} dy = \int {{e^{\alpha x}}} dx$
$ \Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$ ..... (i)
Put (x, y) = (ln2, ln2)
${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$ ..... (ii)
Put (x, y) $ \equiv $ (0, $-$ln2) in (i)
$ - 2 = {1 \over \alpha } + C$ ..... (iii)
(ii) $-$ (iii)
${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$
$\Rightarrow$ $\alpha$ = 2 (as $\alpha$ $\in$ N)
$ \Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$ ..... (i)
Put (x, y) = (ln2, ln2)
${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$ ..... (ii)
Put (x, y) $ \equiv $ (0, $-$ln2) in (i)
$ - 2 = {1 \over \alpha } + C$ ..... (iii)
(ii) $-$ (iii)
${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$
$\Rightarrow$ $\alpha$ = 2 (as $\alpha$ $\in$ N)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
If $y = y(x),y \in \left[ {0,{\pi \over 2}} \right)$ is the solution of the differential equation $\sec y{{dy} \over {dx}} - \sin (x + y) - \sin (x - y) = 0$, with y(0) = 0, then $5y'\left( {{\pi \over 2}} \right)$ is equal to ______________.
Correct Answer: 2
Explanation:
$\sec y{{dy} \over {dx}} = 2\sin x\cos y$
${\sec ^2}ydy = 2\sin xdx$
$\tan y = - 2\cos x + c$
$c = 2$
$\tan y = - 2\cos x + 2 \Rightarrow $ at $x = {\pi \over 2}$
$\tan y = 2$
${\sec ^2}y{{dy} \over {dx}} = 2\sin x$
$ \therefore $ $5{{dy} \over {dx}} = 2$
${\sec ^2}ydy = 2\sin xdx$
$\tan y = - 2\cos x + c$
$c = 2$
$\tan y = - 2\cos x + 2 \Rightarrow $ at $x = {\pi \over 2}$
$\tan y = 2$
${\sec ^2}y{{dy} \over {dx}} = 2\sin x$
$ \therefore $ $5{{dy} \over {dx}} = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
Let a curve y = f(x) pass through the point (2, (loge2)2) and have slope ${{2y} \over {x{{\log }_e}x}}$ for all positive real value of x. Then the value of f(e) is equal to ______________.
Correct Answer: 1
Explanation:
$y' = {{2y} \over {x\ln x}}$
$ \Rightarrow {{dy} \over y} = {{2dx} \over {x\ln x}}$
$ \Rightarrow \ln |y| = 2\ln |\ln x| + C$
put x = 2, y = (ln2)2
$\Rightarrow$ c = 0
$\Rightarrow$ y = (lnx)2
$\Rightarrow$ f(e) = 1
$ \Rightarrow {{dy} \over y} = {{2dx} \over {x\ln x}}$
$ \Rightarrow \ln |y| = 2\ln |\ln x| + C$
put x = 2, y = (ln2)2
$\Rightarrow$ c = 0
$\Rightarrow$ y = (lnx)2
$\Rightarrow$ f(e) = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Let y = y(x) be solution of the following differential equation ${e^y}{{dy} \over {dx}} - 2{e^y}\sin x + \sin x{\cos ^2}x = 0,y\left( {{\pi \over 2}} \right) = 0$ If $y(0) = {\log _e}(\alpha + \beta {e^{ - 2}})$, then $4(\alpha + \beta )$ is equal to ______________.
Correct Answer: 4
Explanation:
Let ey = t
$ \Rightarrow {{dt} \over {dx}} - (2\sin x)t = - \sin x{\cos ^2}x$
$I.F. = {e^{2\cos x}}$
$ \Rightarrow t.{e^{2\cos x}} = \int {{e^{2\cos x}}.( - \sin x{{\cos }^2}x)dx} $
$ \Rightarrow {e^y}.{e^{2\cos x}} = \int {{e^{2z}}.{z^2}dz,z = {e^{2\cos x}}} $
$ \Rightarrow {e^y}.{e^{2\cos x}} = {1 \over 2}.{\cos ^2}x.{e^{2\cos x}} - {1 \over 2}\cos x.{e^{2\cos x}} + {{{e^{2\cos x}}} \over 4} + C$
at $x = {\pi \over 2},y = 0 \Rightarrow C = {3 \over 4}$
$ \Rightarrow {e^y} = {1 \over 2}{\cos ^2}x - {1 \over 2}\cos x + {1 \over 4} + {3 \over 4}.{e^{ - 2\cos x}}$
$ \Rightarrow y = \log \left[ {{{{{\cos }^2}x} \over 2} - {{\cos x} \over 2} + {1 \over 4} + {3 \over 4}{e^{ - 2\cos x}}} \right]$
Put x = 0
$ \Rightarrow y = \log \left[ {{1 \over 4} + {3 \over 4}{e^{ - 2}}} \right] \Rightarrow \alpha = {1 \over 4},\beta = {3 \over 4}$
$ \Rightarrow {{dt} \over {dx}} - (2\sin x)t = - \sin x{\cos ^2}x$
$I.F. = {e^{2\cos x}}$
$ \Rightarrow t.{e^{2\cos x}} = \int {{e^{2\cos x}}.( - \sin x{{\cos }^2}x)dx} $
$ \Rightarrow {e^y}.{e^{2\cos x}} = \int {{e^{2z}}.{z^2}dz,z = {e^{2\cos x}}} $
$ \Rightarrow {e^y}.{e^{2\cos x}} = {1 \over 2}.{\cos ^2}x.{e^{2\cos x}} - {1 \over 2}\cos x.{e^{2\cos x}} + {{{e^{2\cos x}}} \over 4} + C$
at $x = {\pi \over 2},y = 0 \Rightarrow C = {3 \over 4}$
$ \Rightarrow {e^y} = {1 \over 2}{\cos ^2}x - {1 \over 2}\cos x + {1 \over 4} + {3 \over 4}.{e^{ - 2\cos x}}$
$ \Rightarrow y = \log \left[ {{{{{\cos }^2}x} \over 2} - {{\cos x} \over 2} + {1 \over 4} + {3 \over 4}{e^{ - 2\cos x}}} \right]$
Put x = 0
$ \Rightarrow y = \log \left[ {{1 \over 4} + {3 \over 4}{e^{ - 2}}} \right] \Rightarrow \alpha = {1 \over 4},\beta = {3 \over 4}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
Let y = y(x) be the solution of the differential equation $\left( {(x + 2){e^{\left( {{{y + 1} \over {x + 2}}} \right)}} + (y + 1)} \right)dx = (x + 2)dy$, y(1) = 1. If the domain of y = y(x) is an open interval ($\alpha$, $\beta$), then | $\alpha$ + $\beta$| is equal to ______________.
Correct Answer: 4
Explanation:
Let y + 1 = Y and x + 2 = X
dy = dY
dx = dX
$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$
$ \Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$
$ \Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$
$ \Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$
$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$
$\because$ (1, 1) satisfy this equation
So, $c = - {e^{ - {2 \over 3}}} - \ln 3$
Now, $y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$
Domain :
$\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}$
$ \Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}$
$ \Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}$
$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2$
So, $\alpha + \beta = - 4$
$ \Rightarrow \left| {\alpha + \beta } \right| = 4$
dy = dY
dx = dX
$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$
$ \Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$
$ \Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$
$ \Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$
$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$
$\because$ (1, 1) satisfy this equation
So, $c = - {e^{ - {2 \over 3}}} - \ln 3$
Now, $y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$
Domain :
$\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}$
$ \Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}$
$ \Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}$
$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2$
So, $\alpha + \beta = - 4$
$ \Rightarrow \left| {\alpha + \beta } \right| = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
Let a curve y = y(x) be given by the solution of the differential equation $\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$. If it intersects y-axis at y = $-$1, and the intersection point of the curve with x-axis is ($\alpha$, 0), then e$\alpha$ is equal to __________________.
Correct Answer: 2
Explanation:
$\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$
Put cos$-$1(e$-$x) $\theta$, $\theta$ $\in$ [0, $\pi$]
$\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}$
$\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}} $
$\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy$
${1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $
Put ${e^x} = t,{{dt} \over {dx}} = {e^x}$
${1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $
$\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y} $
Put $t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}$
$\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} } $
$ - \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y} $
${{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c$
$2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c$
$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 y + c\buildrel {(0, - 1)} \over \longrightarrow \Rightarrow c = \sqrt 2 $
$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1)$, passes through ($\alpha$, 0)
$2{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2 $
$\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}$
${e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2$
Put cos$-$1(e$-$x) $\theta$, $\theta$ $\in$ [0, $\pi$]
$\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}$
$\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}} $
$\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy$
${1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $
Put ${e^x} = t,{{dt} \over {dx}} = {e^x}$
${1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $
$\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y} $
Put $t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}$
$\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} } $
$ - \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y} $
${{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c$
$2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c$
$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 y + c\buildrel {(0, - 1)} \over \longrightarrow \Rightarrow c = \sqrt 2 $
$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1)$, passes through ($\alpha$, 0)
$2{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2 $
$\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}$
${e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Let y = y(x) be the solution of the differential equation
xdy $-$ ydx = $\sqrt {({x^2} - {y^2})} dx$, x $ \ge $ 1, with y(1) = 0. If the area bounded by the line x = 1, x = e$\pi$, y = 0 and y = y(x) is $\alpha$e2$\pi$ + $\beta$, then the value of 10($\alpha$ + $\beta$) is equal to __________.
xdy $-$ ydx = $\sqrt {({x^2} - {y^2})} dx$, x $ \ge $ 1, with y(1) = 0. If the area bounded by the line x = 1, x = e$\pi$, y = 0 and y = y(x) is $\alpha$e2$\pi$ + $\beta$, then the value of 10($\alpha$ + $\beta$) is equal to __________.
Correct Answer: 4
Explanation:
$xdy - ydx = \sqrt {{x^2} - {y^2}} dx$
dividing both sides by x2, we get
${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$
$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$
$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$
Integrating both side, we get
$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $
${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$
Given, y(1) = 0 $ \Rightarrow $ at x = 1, y = 0
$ \therefore $ $ \Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$
$ \Rightarrow $ C = 0
$ \therefore $ ${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$
$ \Rightarrow $ y = x sin(ln(x))
$ \therefore $ Area $ = \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$
Let, lnx = t
$ \Rightarrow $ x = et
$ \Rightarrow $ dx = et dt
New lower limit, t = ln(1) = 0
and upper limit t = ln$({e^{\pi {} }})$ = ${\pi {} }$
$ \therefore $ Area = $\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$
$ = \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$
$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$
$ = {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$
$ = {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$
$ = \alpha {e^{2\pi {} }} + \beta $
$ \therefore $ $\alpha = {1 \over 5},\beta = {1 \over 5}$
So, $10(\alpha + \beta ) = 4$
dividing both sides by x2, we get
${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$
$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$
$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$
Integrating both side, we get
$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $
${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$
Given, y(1) = 0 $ \Rightarrow $ at x = 1, y = 0
$ \therefore $ $ \Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$
$ \Rightarrow $ C = 0
$ \therefore $ ${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$
$ \Rightarrow $ y = x sin(ln(x))
$ \therefore $ Area $ = \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$
Let, lnx = t
$ \Rightarrow $ x = et
$ \Rightarrow $ dx = et dt
New lower limit, t = ln(1) = 0
and upper limit t = ln$({e^{\pi {} }})$ = ${\pi {} }$
$ \therefore $ Area = $\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$
$ = \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$
$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$
$ = {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$
$ = {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$
$ = \alpha {e^{2\pi {} }} + \beta $
$ \therefore $ $\alpha = {1 \over 5},\beta = {1 \over 5}$
So, $10(\alpha + \beta ) = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Let the curve y = y(x) be the solution of the differential equation, ${{dy} \over {dx}}$ = 2(x + 1). If the numerical value of area bounded by the curve y = y(x) and x-axis is ${{4\sqrt 8 } \over 3}$, then the value of y(1) is equal to _________.
Correct Answer: 2
Explanation:
Given, ${{dy} \over {dx}}$ = 2(x + 1)
Integrating both sides, we get
$y = {x^2} + 2x + c$
Let the two roots of the quadratic equation $\alpha $ and $\beta $
As parabola intercept the x axis so D > 0
From figure, AB = |$\alpha $ - $\beta $| = ${{\sqrt D } \over {\left| a \right|}}$ = $\sqrt D $
and BC = $ - {D \over {4a}}$ = $ - {D \over 4}$
$ \therefore $ Area of rectangle (ABCD) = AB $ \times $ BC = $\sqrt D \times {D \over 4}$
From property we know,
Area of parabola with the x axis = ${2 \over 3}$(Area of rectangle)
$ \Rightarrow $ ${{4\sqrt 8 } \over 3}$ = ${2 \over 3} \times \sqrt D \times {D \over 4}$
$ \Rightarrow $ $D\sqrt D $ = $8\sqrt 8 $
$ \Rightarrow $ D = 8
$ \therefore $ b2 - 4ac = 8
$ \Rightarrow $ 4 - 4c = 8
$ \Rightarrow $ 1 $-$ c = 2 $ \Rightarrow $ c = $-$ 1
Equation of f(x) = x2 + 2x $-$ 1
$ \therefore $ f(1) = 1 + 2 $-$ 1 = 2
Integrating both sides, we get
$y = {x^2} + 2x + c$
Let the two roots of the quadratic equation $\alpha $ and $\beta $
As parabola intercept the x axis so D > 0
From figure, AB = |$\alpha $ - $\beta $| = ${{\sqrt D } \over {\left| a \right|}}$ = $\sqrt D $
and BC = $ - {D \over {4a}}$ = $ - {D \over 4}$
$ \therefore $ Area of rectangle (ABCD) = AB $ \times $ BC = $\sqrt D \times {D \over 4}$
From property we know,
Area of parabola with the x axis = ${2 \over 3}$(Area of rectangle)
$ \Rightarrow $ ${{4\sqrt 8 } \over 3}$ = ${2 \over 3} \times \sqrt D \times {D \over 4}$
$ \Rightarrow $ $D\sqrt D $ = $8\sqrt 8 $
$ \Rightarrow $ D = 8
$ \therefore $ b2 - 4ac = 8
$ \Rightarrow $ 4 - 4c = 8
$ \Rightarrow $ 1 $-$ c = 2 $ \Rightarrow $ c = $-$ 1
Equation of f(x) = x2 + 2x $-$ 1
$ \therefore $ f(1) = 1 + 2 $-$ 1 = 2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
The difference between degree and order of a differential equation that represents the family of curves given by ${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$, a > 0 is _________.
Correct Answer: 2
Explanation:
${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$
Differentiating both sides, we get
$2yy' = a$
${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$
$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$
$y - 2xy' = \sqrt 2 y'\sqrt {yy'} $
${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$
D = 3 & O = 1
$ \therefore $ D $-$ O = 3 $-$ 1 = 2
Differentiating both sides, we get
$2yy' = a$
${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$
$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$
$y - 2xy' = \sqrt 2 y'\sqrt {yy'} $
${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$
D = 3 & O = 1
$ \therefore $ D $-$ O = 3 $-$ 1 = 2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
If y = y(x) is the solution of the equation
${e^{\sin y}}\cos y{{dy} \over {dx}} + {e^{\sin y}}\cos x = \cos x$, y(0) = 0; then
$1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$ is equal to ____________.
${e^{\sin y}}\cos y{{dy} \over {dx}} + {e^{\sin y}}\cos x = \cos x$, y(0) = 0; then
$1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$ is equal to ____________.
Correct Answer: 1
Explanation:
esin y cos y${{dy} \over {dx}}$ + esin y cos x = cos x
Put esin y = t
esin y $\times$ cos y${{dy} \over {dx}}$ = ${{dt} \over {dx}}$
$ \Rightarrow $ ${{dt} \over {dx}}$ + t cos x = cos x
I. F. = ${e^{\int {\cos x\,dx} }} = {e^{\sin x}}$
Solution of differential equation :
$t.{e^{\sin x}} = \int {{e^{\sin x}}.\cos x\,dx} $
${e^{\sin y}}.{e^{\sin x}} = {e^{\sin x}} + c$
at x = 0, y = 0
1 = 1 + c $ \Rightarrow $ c = 0
$ \therefore $ esin x + sin y = esin x
$ \Rightarrow $ sin x + sin y = sin x
$ \Rightarrow $ sin y = 0 $ \Rightarrow $ y = 0
$ \Rightarrow y\left( {{\pi \over 6}} \right) = 0,y\left( {{\pi \over 3}} \right) = 0,y\left( {{\pi \over 4}} \right) = 0$
$ \therefore $ $1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$
= 1 + 0 + 0 + 0 = 1
Put esin y = t
esin y $\times$ cos y${{dy} \over {dx}}$ = ${{dt} \over {dx}}$
$ \Rightarrow $ ${{dt} \over {dx}}$ + t cos x = cos x
I. F. = ${e^{\int {\cos x\,dx} }} = {e^{\sin x}}$
Solution of differential equation :
$t.{e^{\sin x}} = \int {{e^{\sin x}}.\cos x\,dx} $
${e^{\sin y}}.{e^{\sin x}} = {e^{\sin x}} + c$
at x = 0, y = 0
1 = 1 + c $ \Rightarrow $ c = 0
$ \therefore $ esin x + sin y = esin x
$ \Rightarrow $ sin x + sin y = sin x
$ \Rightarrow $ sin y = 0 $ \Rightarrow $ y = 0
$ \Rightarrow y\left( {{\pi \over 6}} \right) = 0,y\left( {{\pi \over 3}} \right) = 0,y\left( {{\pi \over 4}} \right) = 0$
$ \therefore $ $1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$
= 1 + 0 + 0 + 0 = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
If the curve, y = y(x) represented by the solution of the differential equation (2xy2 $-$ y)dx + xdy = 0, passes through the intersection of the lines, 2x $-$ 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to _________.
Correct Answer: 1
Explanation:
Given,
$(2x{y^2} - y)dx + xdx = 0$
$ \Rightarrow {{dy} \over {dx}} + 2{y^2} - {y \over x} = 0$
$ \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} + {1 \over y}\left( {{1 \over x}} \right) = 2$
${1 \over y} = z$
$ - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dz} \over {dx}}$
$ \Rightarrow {{dz} \over {dx}} + z\left( {{1 \over x}} \right) = 2$
I. F. $ = {e^{\int {{1 \over x}dx} }} = x$
$ \therefore $ $z(x) = \int {2(x)dx} = {x^2} + c$
$ \Rightarrow {x \over y} = {x^2} + c$
As it passes through P(2, 1)
[Point of intersection of $2x - 3y = 1$ and $3x + 2y = 8$]
$ \therefore $ ${2 \over 1} = 4 + c$
$ \Rightarrow c = - 2$
$ \Rightarrow {x \over y} = {x^2} - 2$
Put x = 1
${1 \over y} = 1 - 2 = - 1$
$ \Rightarrow y(1) = - 1$
$ \Rightarrow |y(1)|\, = 1$
$(2x{y^2} - y)dx + xdx = 0$
$ \Rightarrow {{dy} \over {dx}} + 2{y^2} - {y \over x} = 0$
$ \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} + {1 \over y}\left( {{1 \over x}} \right) = 2$
${1 \over y} = z$
$ - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dz} \over {dx}}$
$ \Rightarrow {{dz} \over {dx}} + z\left( {{1 \over x}} \right) = 2$
I. F. $ = {e^{\int {{1 \over x}dx} }} = x$
$ \therefore $ $z(x) = \int {2(x)dx} = {x^2} + c$
$ \Rightarrow {x \over y} = {x^2} + c$
As it passes through P(2, 1)
[Point of intersection of $2x - 3y = 1$ and $3x + 2y = 8$]
$ \therefore $ ${2 \over 1} = 4 + c$
$ \Rightarrow c = - 2$
$ \Rightarrow {x \over y} = {x^2} - 2$
Put x = 1
${1 \over y} = 1 - 2 = - 1$
$ \Rightarrow y(1) = - 1$
$ \Rightarrow |y(1)|\, = 1$
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 :
${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$,
$0 < x < {\pi \over 2}$, 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)
$\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 :
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 :
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)
${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$ 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 :
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 :
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 :
(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 :
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 :
${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$, 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 :
$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$, |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 :
$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$, 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
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
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
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 :
${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$, 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 :
${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$, $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$,
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 :-
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 :
$x{{dy} \over {dx}} + 2y$ = 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 :
${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$
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.
${e \over 4}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
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
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
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 $\left( {{1 \over 2},1} \right)$