Differential Equations Questions - JEE

Population $P(t)$ satisfies $\dfrac{dP}{dt} = 0.5P - 450$, $P(0)=850$. When does $P=0$?

Singular solution of \$\\left(\\dfrac{dy}{dx}\\right)^2 - x\\dfrac{dy}{dx} + y = 0\$.

If the curve $y = y(x)$ is the solution of the differential equation $2(x^5 + x^{5/4})\,dy - y(x + x^{9/4})\,dx = 2x^{9/4}\,dx$, $x > 0$, which passes through the point $\left(1, 1 - \frac{4}{3}\log_e 3\right)$, then the value of $y(16)$ is equal to

The solution of x dy/dx = y + 2√(y² - x²) is

The solution of $y = x\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2$ is

Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y$, where the constant of proportionality $k > 0$ depends on the acceleration due to gravity and the geometry of the hole. If $t$ is measured in minutes and $k = \frac{1}{15}$, then the time to drain the tank if the water is 4 meter deep to start with is

The family of curves represented by $\frac{dy}{dx} = \frac{x + y}{x}$ is

The solution of $\frac{dy}{dx} = y - x$ is given by

Spherical rain drops evaporate at a rate proportional to their surface area at any instant $t$. The differential equation for radius $r$ is:

\$(e^y+1)\\cos x\\,dx+e^y\\sin x\\,dy=0\$, \$y(0)=0\$. Find \$1+y(\\pi/6)+\\int_0^{\\pi/6}\\sin x\\,dx\$.

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

\$\\dfrac{dy}{dx}=(y+1)[(y+1)e^{x^2/2}-x]\$, \$y(0)=0\$. Find \$10\\int_0^1 y\\,dx\$ approximately.

\$(x+1)\\dfrac{dy}{dx}-2(x^2+x)y=e^{x^2}\$, \$y(0)=0\$. Find \$y(2)\$.

The solution of the differential equation ydx - xdy + xy^2 dx = 0, is

The degree of the differential equation satisfying the relation$\sqrt{1 + x^2} + \sqrt{1 + y^2} = \lambda\left(x\sqrt{1 + y^2} - y\sqrt{1 + x^2}\right)$ is

The solution of $x^2\,dy - y^2\,dx + xy(x-y)\,dy = 0$ is $\ln\left|\dfrac{x-y}{xy}\right| = \dfrac{y^k}{2} + c$, then the value of $k$ is

Let $y = f(x)$ and $\frac{x}{y}\frac{dy}{dx} = \frac{3x^2 - y}{2y - x^2}$; $f(1) = 1$ then the possible value of $f(3)$ equals:

The curve satisfying $2xy(y^2\cos(x^2y)-1) + x^2y'(y^2\cos(x^2y)+1)=0$ and passing through $(0,1)$ is

\$\\sin x\\,\\dfrac{dy}{dx}+y\\cos x=4x\$, \$y(\\pi/2)=0\$. Find \$y(\\pi/6)\$.

Let $y=y(x)$ satisfy the differential equation $\left(2xy + x^2y + \dfrac{y^3}{3}\right)dx + \left(x^2+y^2\right)dy=0$. If $y(1)=1$ and $(y(0))^3=ke$, $k\in\mathbb{N}$, then $k$ is

Consider the two statements:Statement (I): $y = xf\\!\left(\tfrac{x}{y}\right)$ satisfies the differential equation $x\,\frac{dy}{dx} - y = 0$.Statement (II): $y = xf\\!\left(\tfrac{x}{y}\right)$ is always a homogeneous function of degree 1.The value of the correct statement(s) is:

A curve $C$ passes through the origin with slope $\dfrac{dy}{dx} = \dfrac{y}{x} + \sec\dfrac{y}{x}$. The equation of $C$ is:

A function $y=f(x)$ satisfies $(x+1)f'(x) - 2(x^2+x)f(x)=0$, $f(0)=1$. Let $S = \{x: f(x) > 1\}$. Then $S$ is:

The solution of x^2 dy - y^2 dx + xy^2(x - y)dy = 0, is

Given that the slope of the tangent to a curve $y = y(x)$ at any point $(x, y)$ is $\dfrac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2 + y^2 - 2x - 2y = 0$, then its equation is:

Water is drained from a vertical cylindrical tank. The rate of drop of water level: $\dfrac{dy}{dt} = -k\sqrt{y}$. Given $y(0) = 4$. Find time to drain ($y=0$).

The value of the constants $m$ and $c$ for which $y = mx + c$ is a solution of the differential equation $2y'' - 5y' - 4y = -4x$ is:

The solution of differential equation xdy(y^2 e^{xy} + e^{x/y}) = ydx(e^{x/y} - y^2e^{xy}), is

\$\\dfrac{dy}{dx}+\\dfrac{2xy}{1+x^2}=\\dfrac{2}{(1+x^2)^2}\$, \$y(0)=0\$. Find \$5y(1)\$.

Solution of the differential equation \(\cos x\,dy = y(\sin x - y)\,dx,\; 0

\$\\dfrac{dy}{dx}=\\dfrac{2\\sqrt{y}}{(1-x)\\sqrt{1-x}}\$, \$y(0)=1\$. Find \$y(1/2)\$.

If $\int x\,e^x\,dx = f(x)$ and the solution of the differential equation $\frac{dy}{dx} = 1 + xy\,is\,y = ke^{f(x/2)} + Ce^x$, then the value of $k$ is equal to (where $C$ is the constant of integration)

If the solution of the differential equation $y^2e^{x^2}dx + \sin(x^2)y'dx = \frac{5}{6}dx\,2\sin(x^2)y'e = x^2 + C$ (where $C$ is an arbitrary constant), then the value of $k$ is equal to

The equation of the family of curves shown in the figure with one arbitrary constant.

Given $\dfrac{dy}{dx} + \left(\dfrac{2x+1}{x}\right)y = e^{-2x}$, and the curve passes through $\left(1,\,\dfrac{1}{2}e^{-2}\right)$. Find $y(\log_e 2)$.

If $(2xy-y^2-y)dx=(2xy+x-x^2)dy$ and $y(1)=1$, then the value of $12|y(-1)|$ is

If $(2x+y^2)\,dx + (2y+3x^2)\,dy = 0$ is exact, the solution is:

Let $\frac{x\,dy}{dx} - y = x^2\left(xe^x + e^x - 1\right)$ for all $x \in \mathbb{R} - \{0\}$ such that $y(1) = e - 1$. If $y(2) = k\,y(1)\,(y(1) + 2)$, then the value of $\dfrac{k^2}{5}$ is

The solution of the differential equation $x\,dy + y\,dx = 0$ passes through the point $(2, 8)$. The latus rectum of the conic represented by the solution curve equals

\$\\dfrac{dy}{dx}-\\dfrac{y+3}{x+2}=0\$. The family of solutions is:

Let $y=y(x)$ be the solution of the differential equation $(1+y^2)e^{\tan x}\,dx+\cos^2x(1+e^{2\tan x})\,dy=0$, $y(0)=1$. Then $y\left(\dfrac{\pi}{4}\right)$ is equal to:

Let $y=y(x)$ be the solution curve of the differential equation $\sec y\dfrac{dy}{dx}+2x\sin y=x^3\cos y$, $y(1)=0$. Then $y(\sqrt{3})$ is equal to:

Let $\alpha|x|=|y|e^{xy-\beta}$, $\alpha,\beta\in\mathbb{N}$ be the solution of the differential equation $x\,dy-y\,dx+xy(x\,dy+y\,dx)=0$, $y(1)=2$. Then $\alpha+\beta$ is equal to ________.

The solution curve of the differential equation $2y\dfrac{dy}{dx}+3=5\dfrac{dy}{dx}$, passing through the point $(0,1)$ is a conic, whose vertex lies on the line:

The solution of $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ is:

If the solution curve of $(y-2\ln x)\,dx+(x\ln x^2)\,dy=0$, $x>1$ passes through $(e,\frac{4}{3})$ and $(e^4,\alpha)$, then $\alpha$ is equal to _______.

Let $y=y(x)$ be a solution of $(1-x^2y^2)\,dx=y\,dx+x\,dy$. If $x=1$ gives $y=2$ and $x=2$ gives $y=\alpha$, then a value of $\alpha$ is

If $y=y(x)$ is the solution of $\dfrac{dy}{dx}+\dfrac{4x}{x^2-1}y=\dfrac{x+2}{5(x^2-1)^2}$, $x>1$, with $y(2)=\dfrac{2}{9}\ln_e(2+\sqrt{3})$, and $y(\sqrt{2})=\alpha\ln_e(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}$, then $\alpha\beta\gamma$ is equal to

Let $y=y(x)$, $y>0$, be a solution of $(1+x^2)\,dy=y(x-y)\,dx$ with $y(0)=1$ and $y(2\sqrt{2})=\beta$. Then

99. 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:

Differential Equations Questions (544)