Differential Equations
419 Questions
2022
TS-EAMCET
MCQ
TS EAMCET 2022 (Online) 18th July Morning Shift
The degree of the differential equation
$ x\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{5 / 3}+7 \frac{d y}{d x}+y=0 $
A.
15
B.
5
C.
12
D.
3
2022
TS-EAMCET
MCQ
TS EAMCET 2022 (Online) 18th July Morning Shift
The curve that satisfies the differential equation $x y d y-\left(1+y^2\right) d x=0$ passes through $(1,0)$ and intersects the curve $x^2+3 y^2=3$ at an angle $\theta$. Then, $\frac{2 \theta}{\pi}=$
A.
2
B.
0
C.
4
D.
1
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
The general solution of the differential equation $\frac{d y}{d x}=\cos ^2(3 x+y)$ is $\tan ^{-1}\left(\frac{\sqrt{3}}{2} \tan (3 x+y)\right)=f(x)$. Then, $f(x)=$
A.
$2 \sqrt{3}(x+C)$
B.
$x+C$
C.
$\frac{x+C}{2 \sqrt{3}}$
D.
$\frac{\sqrt{3}}{2}(x+C)$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
If the general solution of the differential equation $\cos ^2 x \frac{d y}{d x}+y=\tan x$ is $y=\tan x-1+C e^{-\tan x}$ satisfies $y\left(\frac{\pi}{4}\right)=1$, then $C=$
A.
e
B.
1
C.
$-$1
D.
$\frac{1}{e}$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
Assertion (A) Order of the differential equations of a family of circles with constant radius is two.
Reason (R) An algebraic equation having two arbitrary constants is general solution of a second order differential equation.
A.
A and R are true, R is the correct explanation to A
B.
A is true, R is false
C.
A and R are true, R is not the correct explanation to A
D.
A is false, R is true
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
If $l$ and $m$ are order and degree of a differential equation of all the straight lines at constant distance of $P$ units from the origin, then $l m^2+l^2 m=$
A.
2
B.
6
C.
12
D.
30
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
If $2 x-y+C \log (|x-2 y-4|)=k$ is the general solution of $\frac{d y}{d x}=\frac{2 x-4 y-5}{x-2 y+2}$, then $C=$
A.
4
B.
2
C.
3
D.
$-$4
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
By eliminating the arbitrary constants from $y=(a+b) \sin (x+c)-d e^{x+e+f}$, then differential equation has order of
A.
6
B.
4
C.
3
D.
5
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
If the solution of $\frac{d y}{d x}-y \log _e 0.5=0, y(0)=1$, and $y(x) \rightarrow k$, as $x \rightarrow \infty$, then $k=$
A.
$\infty$
B.
$-1$
C.
1
D.
0
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
$y=A e^x+B e^{-2 x}$ satisfies which of the following differential equations?
A.
$\frac{d^2 y}{d x^2}-\frac{d y}{d x}+2 y=0$
B.
$\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}-y=0$
C.
$\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+y=0$
D.
$\frac{d^2 y}{d x^2}+\frac{d y}{d x}-2 y=0$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
If y = y(x) is the solution curve of the differential equation ${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$ ; x > 0 and y(1) = 1, then $y\left( {{1 \over 2}} \right)$ is equal to :
A.
${3 \over 2} - {1 \over {\sqrt e }}$
B.
$3 + {1 \over {\sqrt e }}$
C.
3 + e
D.
3 $-$ e
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
If ${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$, y(0) = 0, then for y = 1, the value of x lies in the interval :
A.
(1, 2)
B.
$\left( {{1 \over 2},1} \right]$
C.
(2, 3)
D.
$\left( {0,{1 \over 2}} \right]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
If $y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$, x > 0, $\phi$ > 0, and y(1) = $-$1, then $\phi \left( {{{{y^2}} \over 4}} \right)$ is equal to :
A.
4 $\phi$ (2)
B.
4$\phi$ (1)
C.
2 $\phi$ (1)
D.
$\phi$ (1)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
If ${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$, y(0) = 1, then y(1) is equal to :
A.
log2(2 + e)
B.
log2(1 + e)
C.
log2(2e)
D.
log2(1 + e2)
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 :
${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$ 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 :
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 :
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 :
${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$, 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 :
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 :
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 :
${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$, 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 :
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