Differential Equations Questions - JEE

Question 21: If $\frac{f'(x)}{f(x)} = 1$ and $f(1) = 2$, then find $\left[\frac{f(3)}{2}\right]$, where $[\cdot]$ denotes the greatest integer function.

If the solution $y(x)$ of the given differential equation $(e^y+1)\cos x\,dx+e^y\sin x\,dy=0$ passes through the point $\left(\dfrac{\pi}{2},0\right)$, then the value of $e^{y(\pi/6)}$ is equal to ________.

The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y\frac{dy}{dx} + (x^2 - y)\frac{dy}{dx} - x^2 = 0$ can be

Let $y(x)$ be the solution of the differential equation $(x\log x)\frac{dy}{dx} + y = 2x\log x,\ (x \geq 1)$. Then $y(e)$ is equal to

Use the methods of solving first order differential equation to find the general solution of $\frac{d^2y}{dx^2} = 3\left(\frac{dy}{dx}\right)^{3/2}$.

If $m$ and $n$ are order and degree of the equation $2\left(\frac{d^2y}{dx^2}\right)^4 + 3\frac{d^2y}{dx^2}\left(\frac{d^3y}{dx^3}\right)^5 = x^2 - 1$, then

The solution of the differential equation $\frac{dy}{dx} = \frac{y - x + 1}{y + x + 5}$ is

Let $f$ be a continuous function satisfying the equation $\displaystyle\int_0^x f(t)\,dt + \int_0^x (x-t)\cdot f(t)\,dt = e^{-x} - 1$, then

Let $y = f(x)$ be a curve passing through $(e, e^e)$, which satisfies the differential equation $(2ny + xy \log_e x)\,dx - x\log_e x\,dy = 0$, $x > 0, y > 0$. If $g(x) = \lim_{n \to \infty} f(x)$, then $\int_{1/e}^{e} g(x)\,dx$ equals

If $y + \dfrac{d}{dx}(xy) = x(\sin x + \log x)$, then

The solution of the differential equation $\sin^2 y \frac{dy}{dx} + 2\tan x \cos^2 y = 2\sec x \cos^3 y$ is:

The solution of $(2x - 10y^3)\frac{dy}{dx} + y = 0$ is $xy^2 + ly^5 + C$. Then $l$ is

The order of the differential equation whose general solution is given by $y = (c_1 + c_2x)\cos(x + c_3) - c_4e^{-7x}$, where $c_1, c_2, c_3, c_4$ & $c_5$ are arbitrary constants, is

We have, $\frac{dy}{dx} + \left(\frac{-1}{x}\right)y = x\left(xe^x + e^x - 1\right)$. If $y(x=1) = e-1$, find $k$ such that $y(2) = k \cdot y(1) \cdot [y(1)+2]$, and compute $\frac{k^2}{5}$.

The solution of the differential equation \ \int\left(\dfrac{y+1}{y}\right)dy = \int e^x(\sin 2x - \cos^2 x)\,dx \ with \ x=0, y=1 \ gives the particular solution. Find the value of \ C:

The solution of $\frac{dy}{dx} = \frac{xe^{(n-1)y}}{x}$ is

The solution of the differential equation $(1+2e^{x/y})dx + 2e^{x/y}\left(1 - \dfrac{x}{y}\right)dy = 0$ is $(x + 2ye^{x/y}) = c \Rightarrow l - 2$. Find $l$.

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). The equation of the curve given that it passes through (-2, 1) is

The orthogonal trajectories of the family of curves $a^{n-1}y = x^n$ are given by (a is the arbitrary constant)

Let $y = f(x)$ satisfy the differential equation $xy(1 + y)dx = dy$. If $f(0) = 1$ and $f(2) = \frac{e^2}{k - e^2}$, then find the value of k.

The solution of the differential equation $e^{-x}(y+1)\,dy + (\cos^2 x - \sin 2x)y\,dx = 0$ subjected to condition that $y = 1$ when $x = 0$, is:

If $y = f(x)$ satisfies the differential equation $\sin x \dfrac{dy}{dx} + 2y\cos x = 8$ with $f(\pi/2) = 8$, then the minimum value of $f(x)$ is:

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $y(0) = 1$. Then (JEE Main 2019)

A curve is such that the portion of the X-axis cut off between the origin and tangent at a point is twice the abscissa and which passes through the point $(1, 2)$. The equation of the curve is

The solution of the differential equation $\dfrac{dy}{dx}=(4x+y+1)^2$ is

The solution of the differential equation $(1 + y^2) + (x - e^{\tan^{-1}y})\dfrac{dy}{dx} = 0$, is

If $y = f(x)$ satisfy the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^2$; $f(1) = 1$; then value of $f(3)$ equals:

If $y_1(x)$ and $y_2(x)$ are two solutions of $\frac{dy}{dx} + f(x)y = r(x)$, then $y_1(x) + y_2(x)$ is solution of

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

Let the tangent at any point $P$ on a curve passing through $(1,1)$ and $(\frac{1}{10},100)$, intersect positive $x$ and $y$-axes at $A$ and $B$ respectively. If $PA:PB=1:k$ and $y=y(x)$ is the solution of $e^{dy/dx}=kx+\frac{k}{2}$, $y(0)=k$, then $4y(1)-5\ln 3$ is equal to _______.

The solution of $(y(1 + x^{-1}) + \sin y) dx + (x + \ln x + x \cos y) dy = 0$ is

The solution of the equation $\dfrac{d^2y}{dx^2} = e^{-2x}$ is

Let $f(x)$ be a polynomial function satisfying $f'(x) + f(x) = x$. Then the value of $f(4)$ is equal to:

Solution of the differential equation $e^{x^2}(y^2 - 1)dy + e^{y^2} - y dy + e^{x^2}(xy^2 - x) = 0$ is

A function $y = f(x)$ has a second order derivative $f''(x) = 6(x - 1)$. If its graph passes through the point (2,1) and at the point the tangent to the graph is $y = 3x - 5$, then the function is

If $y=y(x)$ is the solution curve of the differential equation $(x^2-4)\,dy-(y^2-3y)\,dx=0$, $x>2$, $y(4)=\dfrac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals:

The solution of $\frac{dy}{dx}(x^2y^3 + xy) = 1$ is

The solution of $\frac{dy}{dx} = (x + y - 1)^2 + \frac{y}{x - 1}$ is $\ln(x + y) = 0$

Let $x=x(y)$, $0<y<\frac{\pi}{2}$, satisfy $(\ln\cos y)^2\cos y\,dx-(1+3x\ln\cos y)\sin y\,dy=0$ with $x(\frac{\pi}{3})=\frac{1}{2\ln 2}$. If $x(\frac{\pi}{6})=\frac{1}{\ln m-\ln n}$, then $mn$ is equal to

If the solution of the differential equation is $Ax^2 + By^2 = 1$, then the order and degree of the resulting differential equation (after eliminating $A$ and $B$) are:

\$(3x^2y+y^3)\\,dx+(x^3+3xy^2)\\,dy=0\$. General solution:

Order and degree of \$\\left[1+\\left(\\dfrac{dy}{dx}\\right)^2\\right]=\\dfrac{d^2y}{dx^2}\$ are respectively:

If $y(t)$ is a solution of $(1 + t)\frac{dy}{dt} - ty = 1$ and $y(0) = -1$, then $y(1)$ = _____.

A function $f(x)$ satisfies $\displaystyle\int_0^x f(t)\,dt = n\,f(x)$ for all $x > 0$. Which is true?

The solution of \frac{xdy}{x^2 + y^2} - \frac{y}{x^2 + y^2}dx, is

The solution of the differential equation \ \dfrac{dx}{dy} + \dfrac{x^2}{y^2} - \dfrac{x}{y} + 1 = 0 \ is:

\$y\\,dx-(x+2y^2)\\,dy=0\$ has general solution:

The solution of the differential equation $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x}\cos y \sin y$, where $y = -1$ as $x \to \infty$, is

Let $f$ be differentiable with $x^2f(x)-x=4\displaystyle\int_0^x tf(t)\,dt$, $f(1)=\dfrac{2}{3}$. Then $18f(3)$ is equal to

\$(x^2-y^2)\\,dx+2xy\\,dy=0\$ through \$(1,1)\$. Find the curve.

Differential Equations Questions (544)