JEE Mains JEE Advance BITSAT
JEE Mains | PYQ

Year

All years 2026 2025 2024 2023 2022 2021

Let \( y = y_1(x) \) and \( y = y_2(x) \) be the solution curves of the differential equation \( \dfrac{dy}{dx} = y + 7 \) with initial conditions \( y_1(0) = 0 \) and \( y_2(0) = 1 \) respectively. Then the curves \( y = y_1(x) \) and \( y = y_2(x) \) intersect at.

Let \( y = y(x) \) be the solution of the differential equation \(\displaystyle \frac{dy}{dx} + \frac{5}{x(x^5 + 1)}y = \frac{(x^5 + 1)^2}{x^7} , \; x>0 \). If \( y(1) = 2 \), then \( y(2) \) is equal to:

Let a curve \( y=f(x), \, x \in (0,\infty) \) pass through the points \( P\!\left(1,\dfrac{3}{2}\right) \) and \( Q\!\left(a,\dfrac{1}{2}\right) \). If the tangent at any point \( R(b,f(b)) \) to the given curve cuts the \( y \)-axis at the point \( S(0,c) \) such that \( bc=3 \), then \( (PQ)^2 \) is equal to ________ .

Let \( y = y(x) \) be a solution curve of the differential equation, \( (1 - x^2y^2)\,dx = y\,dx + x\,dy \). If the line \( x = 1 \) intersects the curve \( y = y(x) \) at \( y = 2 \) and the line \( x = 2 \) intersects the curve \( y = y(x) \) at \( y = \alpha \), then a value of \( \alpha \) is:

Let the solution curve \( y=y(x) \) of the differential equation\( \dfrac{dy}{dx}+\left(\dfrac{-3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}\right)y=2x\,exp{\left\{\dfrac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right\}}\) pass through origin. Then \( y(1) \) is equal to:

Let \( y = f(x) \) be the solution of the differential equation \( y(x+1)dx - x^2 dy = 0 \), \( y(1) = e \). Then \( \displaystyle \lim_{x \to 0^+} f(x) \) is equal to:

Let \( y=y(x) \) be the solution of the differential equation \( x^{3}dy+(xy-1)dx=0,\, x>0,\, y\!\left(\dfrac{1}{2}\right)=3-e \). Then \( y(1) \) is equal to:

Let \( y = y(x) \) be the solution of the differential equation \( (x^{2} - 3y^{2})\,dx + 3xy\,dy = 0, \, y(1) = 1 \). Then \( 6y^{2}(e) \) is equal to:

If \( f:\mathbb{R} \to \mathbb{R} \) be a differentiable function such that \( f'(x) + f(x) = \int_{0}^{2} f(t) \, dt \). If \( f(0) = e^{-2} \), then \( 2f(0) - f(2) \) is equal to __________.

Let \( f \) be a differentiable function such that \(\displaystyle x^{2}f(x)-x = 4\!\int_{0}^{x} t f(t)\,dt \), \( f(1)=\dfrac{2}{3} \). Then \( 18f(3) \) is equal to:

Let \( y = y(x) \) be the solution of the differential equation \( (3y^{2} - 5x^{2})y \, dx + 2x(x^{2} - y^{2}) \, dy = 0 \) such that \( y(1) = 1 \). Then \( \left| \left(y(2)\right)^{-3} - 12y(2) \right| \) is equal to:

The solution of the differential equation \( \frac{dy}{dx}=\left(\frac{x^{2}+3y^{2}}{3x^{2}+y^{2}}\right) \), \( y(1)=0 \) is:

Let \( y = y(x) \) be a solution of the differential equation \( (x\cos x) \, dy + (xy\sin x + y\cos x - 1)\, dx = 0 \), \( 0 < x < \dfrac{\pi}{2} \). If \( \dfrac{\pi}{3} \, y\!\left(\dfrac{\pi}{3}\right) = \sqrt{3} \), then \( \left| \dfrac{\pi}{6}\, y''\!\left(\dfrac{\pi}{6}\right) + 2\, y'\!\left(\dfrac{\pi}{6}\right) \right| \) is equal to __________.

Let \( \alpha x=\exp \left(x^{\beta} y^{\gamma}\right) \) be the solution of the differential equation \( 2x^{2}y\,dy-(1-xy^{2})\,dx=0, \ x>0, \ y(2)=\sqrt{\log_{e}2}. \) Then \( \alpha+\beta-\gamma \) equals:

Let \( x=x(y) \) be the solution of the differential equation \( 2(y+2)\log_{e}(y+2)\,dx+(x+4-2\log_{e}(y+2))\,dy=0,\ y>-1 \) with \( x\!\left(e^{4}-2\right)=1 \). Then \( x\!\left(e^{9}-2\right) \) is equal to:

If the solution curve of the differential equation \( (y-2\log_e x)\,dx+(x\log_e x^2)\,dy=0,\; x>1 \) passes through the points \( \left(e,\dfrac{4}{3}\right) \) and \( (e^4,\alpha) \), then \( \alpha \) is equal to ________.

Let the solution curve \( x=x(y) \), \( 0<y<\dfrac{\pi}{2}\), of the differential equation \((\log_e(\cos y))^2 \cos y \, dx - (1 + 3x \log_e(\cos y)) \sin y \, dy = 0\) satisfy \(x\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2\log_e 2}\).If \(\displaystyle x\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\log_e m - \log_e n}\),where \(m\) and \(n\) are coprime, then \(mn\) is equal to:

If the solution curve \( f(x,y)=0 \) of the differential equation \( (1+\log_{e}x)\dfrac{dx}{dy}-x\log_{e}x=e^{y},\ x>0 \) passes through the points \( (1,0) \) and \( (\alpha,2) \), then \( \alpha^{\alpha} \) is equal to:

Let \( y=y(t) \) be a solution of the differential equation \( \frac{dy}{dt}+\alpha y=\gamma e^{-\beta t} \). Where, \( \alpha>0 \), \( \beta>0 \) and \( \gamma>0 \). Then \( \displaystyle \lim_{t\to\infty} y(t) \).

If \( y=y(x) \) is the solution curve of the differential equation \( \dfrac{dy}{dx}+y\tan x = x\sec x \), \( 0 \le x \le \dfrac{\pi}{3} \), \( y(0)=1 \)
Then \( y\!\left(\dfrac{\pi}{6}\right) \) is equal to:

Let the tangent at any point \( P \) on a curve passing through the points \( (1, 1) \) and \( \left(\dfrac{1}{10}, 100\right) \), intersect positive \( x \)-axis and \( y \)-axis at the points \( A \) and \( B \) respectively. If \( PA : PB = 1 : k \) and \( y(x) \) is the solution of the differential equation \( e^{\dfrac{dy}{dx}} = kx + \dfrac{k}{2},\ y(0) = k, \) then \( 4y(1) - 4\log_{e} 3 \) is equal to ______.

Let \( y = y(x) \) be the solution curve of the differential equation \( \dfrac{dy}{dx} = \dfrac{y}{x}\big(1 + xy^{2}(1 + \log_{e} x)\big) \), \( x > 0 \), \( y(1) = 3 \). Then \( \dfrac{y^{2}(x)}{9} \) is equal to:

Let \( y = y(x) \) be the solution of the differential equation \( x \log_e x \, \frac{dy}{dx} + y = x^2 \log_e x, \, (x > 1) \). If \( y(2) = 2 \), then \( y(e) \) is equal to:

Let \( y=y(x),\ y>0 \), be a solution curve of the differential equation \( (1+x^{2})\,dy=y(x-y)\,dx \). If \( (0)=1 \) and \( y(2\sqrt{2})=\beta \), then:

The slope of tangent at any point \( (x,y) \) on a curve \( y=y(x) \) is \( \dfrac{x^2+y^2}{2xy} \), \( x>0 \). If \( y(2)=0 \), then a value of \( y(8) \) is:

← PreviousQuestion 19 of 25Next →

JEE Main 2023 25th Jan Evening Shift

Let \( y=y(t) \) be a solution of the differential equation \( \frac{dy}{dt}+\alpha y=\gamma e^{-\beta t} \). Where, \( \alpha>0 \), \( \beta>0 \) and \( \gamma>0 \). Then \( \displaystyle \lim_{t\to\infty} y(t) \).