Differential Equations Questions (544)

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 \(k = \dfrac{1}{15}\) then the time (in hour) to drain the tank if the water is 4m deep to start with is

If $\dfrac{dx}{dy}=\dfrac{1+x-y^2}{y}$, $x(1)=1$, then $5x(2)$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\dfrac{dy}{dx}=2x(x+y)^3-x(x+y)-1$, $y(0)=1$. Then $\left(\dfrac{1}{\sqrt{2}}+y\left(\dfrac{1}{\sqrt{2}}\right)\right)^2$ equals:

A function $y=f(x)$ satisfies $f(x)\sin2x+\sin x-(1+\cos^2x)f'(x)=0$ with condition $f(0)=0$. Then $f\left(\dfrac{\pi}{2}\right)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $(1-x^2)\,dy=\left[xy+(x^3+2)\sqrt{3(1-x^2)}\right]\,dx$, $-1<x<1$, $y(0)=0$. If $y\left(\dfrac{1}{2}\right)=\dfrac{m}{n}$, $m$ and $n$ are coprime numbers, then $m+n$ is equal to

For what \\(\\lambda\\) does \\((3xy+y^2)\\,dx+(x^2+\\lambda xy)\\,dy=0\\) have IF \\(=x^n\\)?

The temperature $T(t)$ of a body at time $t=0$ is $160^\circ F$ and it decreases continuously as per the differential equation $\dfrac{dT}{dt}=-K(T-80)$, where $K$ is positive constant. If $T(15)=120^\circ F$, then $T(45)$ is equal to

Let $y = y(x)$ be the solution of the differential equation $\cos x\,(\log_e(\cos x))^2\,dy + (\sin x - 3y\sin x\log_e(\cos x))\,dx = 0$, $x\in\left(0,\dfrac{\pi}{2}\right)$. If $y\!\left(\dfrac{\pi}{4}\right) = \dfrac{-1}{\log_e 2}$, then $y\!\left(\dfrac{\pi}{6}\right)$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\dfrac{dy}{dx}+\dfrac{2x}{(1+x^2)^2}y=xe^{\frac{1}{1+x^2}}$; $y(0)=0$. Then the area enclosed by the curve $f(x)=y(x)e^{-\frac{1}{1+x^2}}$ and the line $y-x=4$ is ________.

Solve xdx + ydy + (xdy - ydx)/(x² + y²) = 0.

Let u(x) and v(x) satisfy the differential equations du/dx + p(x)u = f(x) and dv/dx + p(x)v = g(x) respectively, where p(x), f(x) and g(x) are continuous functions. If u(x₁) > v(x₁) for some x₁ and f(x) > g(x) for all x > x₁, prove that any point (x, y), where x > x₁, does not satisfy the equations y = u(x) and y = v(x).

Depreciation: \(\dfrac{df}{dt} = -k(T-t)\), \(f(0) = \) purchase price \(f_0\), \(f(T)=V(T)\). Find \(k\):

The curve satisfying the differential equation, \(y\,dx - (x + 3y^2)dy = 0\) and passing through the point \((1, 1)\) also passes through the point

Order of ODE with general solution \\(y=c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-2x}+c_5\\).

Let $y = y(x)$ be the solution of the differential equation $\cos x (\log_e(\cos x))^2 dy + (\sin x - 3y\sin x \log_e(\cos x)) dx = 0$, $x \in (0, \pi/2)$. If $y(\pi/4) = \frac{\log_e 2}{1}$, then $y(\pi/6)$ is equal to:

$y = \sin x + \frac{1}{2}\cos ec x$

\\(\\dfrac{dy}{dx}+\\dfrac{2y}{1+x^2}=\\dfrac{4x}{(1+x^2)^2}\\), \\(y(0)=0\\). Find \\(y(1)\\).

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

Solution of \\(\\dfrac{dy}{dx}=\\dfrac{y}{x}+\\tan\\dfrac{y}{x}\\) through \\(\\left(1,\\dfrac{\\pi}{4}\\right)\\) is \\(\\sin\\dfrac{y}{x}=kx\\). Find \\(k\\).

\\(\\dfrac{dy}{dx}=\\dfrac{2y}{x}\\), \\(y(1)=1\\). The curve is:

Area enclosed between \\(y=xe^{-x}\\) and \\(x\\)-axis for \\(x\\in[0,\\infty)\\). Find area.

\\(\\dfrac{dy}{dx}=y\\tan x-2\\sin x\\), \\(y(0)=1\\). Find \\(y(\\pi/3)\\).

If order \\(=m\\) and degree \\(=n\\) for \\(\\left(\\dfrac{d^2y}{dx^2}\\right)^3+\\left(\\dfrac{dy}{dx}\\right)^4+y^5=0\\), find \\(m+n\\).

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

\\((1+x^2)\\dfrac{dy}{dx}+2xy=4x^2\\), \\(y(0)=0\\). Find \\(y(1)\\).

\\(f(xy)=f(x)f(y)\\), \\(f(0)\\ne0\\). \\(y'=f(x)\\), \\(y(0)=1\\). Find \\(y(1/4)+y(3/4)\\).

The singular solution of the differential equation given in previous problem is :

Non-singular solution of the differential equation $x\frac{dy}{dx} = x - \left(\frac{dy}{dx}\right)^2$ is :

If the solution of the differential equation $\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$ be $f(x,y) + \sqrt{1+f(x,y)} = c\left(\sqrt{\frac{x+y}{\sqrt{f(x,y)}}}\right)$, then $f(x,y)$ is:

The solution of the equation $\int_0^x y(t)dt = (x+1)\int_0^x ty(t)dt, x > 0$ as $y = f(x)$ is:

The real value of $m$ for which the substitution $y = u^m$ will transform the differential equation $2x^3y\frac{dy}{dx} + y^4 = 4x^6$ into a homogeneous equation is:

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

The general equation of the equation $y = px + \log p$ which does not contain the singular solution, is :

The solution of $\frac{xdx + ydy}{xdy - ydx} = \frac{a^2 - x^2 - y^2}{x^2 + y^2}$ is:

A curve $f(x)$ passes through the point $P(1,1)$. The normal to the curve at point $P$ is $a(y-1) + (x-1) = 0$. If the slope of the tangent at any point on the curve is proportional to the ordinate at that point, then the equation of the curve is

Given the differential equation $\frac{dy}{dx} = \frac{6x^2}{2y + \cos y}$, $y(1) = \pi$ and the following statements

The equation of the curve satisfying the differential equation $y_2(x^2 + 1) = 2xy_1$ passing through the point (0, 1) and having slope of tangent at $x = 0$ as 3(where $y_2$ and $y_1$ represents $2^{nd}$ and $1^{st}$ order derivative), then:

For the central conics having their axes along the coordinates:

A right circular cylinder with radius $R$ and height $H$ contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = $K > 0$). If $T$ is time after which cylinder will be empty, then

For the differential equation $(3x + 2y^2)dx + 2x(2x + 3y^2)dy = 0$

Let $S_1 = x^2 + y^2 - kx = 0$ and $S_2 = x^2 - y^2 - cx = 0$, then

Consider the differential equation $\cos^2 x \frac{dy}{dx} - (\tan 2x)y = \cos^4 x$; $|x| < \frac{\pi}{4}$ and $y\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{8}$ then:

If a curve $y = f(x)$, passing through the point $(2, 1)$ satisfies the condition that length of subtangent is equal to slope of tangent in $1^{st}$ quadrant given that $\frac{dy}{dx} > 0$, then:

If $f(x)$ is a function such that $\int_0^x (1-t)f(t)dt = \int_0^x f(t)dt; f(1) = 1$, then:

If $y = e^{-x}\sin x$ and $y_n + a_ny_{n-1} = 0$ where $a_n$ is constant for $n \in \mathbb{N}$ & $y_n = \frac{d^ny}{dx^n}$ (nth derivative of $y$), then:

If $y = f(x); f'(x) \geq 0$ & $f(0) = 0$ bounding a curvilinear trapezoid with base $[0, x]$ whose area is proportional to $3^{rd}$ power of $f(x)$. If $f(1) = 3$, then:

The solutions of $y = x\left(\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^3\right)$ are given by (where $p = \frac{dy}{dx}$ and $k$ is constant)

If a tangent drawn to the curve $y = f(x)$ at $(x, y)$ cuts the $x-$ axis and $y-$ axis at $A$ and $B$ respectively such that $\frac{BP}{AP} = \frac{3}{1}$ given $f(1) = 1$, then:

If right circular cone with radius 18 & height 27 contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = $k > 0$). If volume of liquid is $V$ & $r$ is radius of surface of liquid left, then:

A differentiable function satisfies $f(x) = \int_0^x [f(t)\cos t - \cos(t-x)]dt$. which is of the following hold good?