Differential Equations Questions - JEE

Let $y=y(x)$ satisfies the differential equation $y'=\ln(xy'-y)$. If $y(1)=-1$ where $y$ is twice differentiable and $y''(x)\neq 0$, then $y(e)$ equals:

Given the differential equation $\dfrac{dy}{dx} = (x - y)^2$, find the general solution.

The differential equation representing the family of curves $y^2 = 2c(x + \sqrt{c})$, where $c > 0$, is a parameter, is of order and degree as follows:

Solve $x\left(\frac{dy}{dx}\right)^2 + (y - x)\frac{dy}{dx} - y = 0$.

The order and degree of the differential equation $\left(1 + 3\dfrac{dy}{dx}\right)^{2/3} = 4\left(\dfrac{d^3y}{dx^3}\right)$ are respectively:

The solution of the differential equation $x\dfrac{dy}{dx} = y(\log y - \log x + 1)$ is:

If $(2 + \sin x)\dfrac{dy}{dx} + (y+1)\cos x = 0$ and $y(0) = 1$, then $y\!\left(\dfrac{\pi}{2}\right)$ is equal to

Let $y = y(x)$ satisfies the differential equation $y' = \ln(xy' - y)$. If $y(1) = -1$ where $y$ is twice differentiable and $y''(x) \neq 0$, then $y(e)$ equals:

Let $y$ be the solution of the differential equation $\frac{dy}{dx} = -2x(y-1)$ with $f(0) = 1$, then $\lim_{x \to \infty} f(x)$ is equal to

The solution of the differential equation $\sin(x + y)dy = dx$ is

Solve $1 + \left(\frac{dy}{dx}\right)^2 = x\frac{dy}{dx}$.

We have $ \cos x\, dy = y(\sin x - y)\, dx $. The solution is:

We have $y = ae^{3x} + be^x$. Eliminating $a$ and $b$, we get a differential equation of the form $\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 3y = 0$ with $a = 1,\, b = -4,\, c = 3$. Find $a + b + c$.

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

Given $ x\dfrac{dy}{dx} + 2y = x^2 $If $ y(1) = 1 $, find $ y\!\left(\dfrac{1}{2}\right) $.

If $y = (x + \sqrt{1 + x^2})^n$, then the differential equation satisfied by $y$ is:

The differential equation of the family of parabolas having their axis as x-axis is $y^2 = 4a(x - h)$. The order and degree of this differential equation are:

The differential equation $\dfrac{d^2x}{dy^2} + y + \cot^2 x = 0$ must be satisfied by $y = f(x)$ then $f(x)$ may be-

The differential equation $ y(1 + xy)\, dx = x\, dy $ passes through $ (1, -1) $. Find $ f\!\left(-\dfrac{1}{2}\right) $.

Find the differential equation of the family of curves $y = e^x(A\cos x + B\sin x)$ where $A, B$ are arbitrary constants. Also write its order and degree.

The equation of family of circles passing through the origin is $(x-0)^2 + (y-a)^2 = a^2$. The differential equation of this family is:

The general solution of the differential equation $(y^2 - x^3)\,dx - xy\,dy = 0$ $(x \neq 0)$ is (where $c$ is a constant of integration):

The curve which satisfies the differential equation $\tan y + (1+x^2)\cot^{-1}x\left(\dfrac{dy}{dx}\right) = 0$ and passes through $\left(1, \dfrac{\pi}{2}\right)$ is

$b^2 + y^2 - 2y + z + y = c$ and $xdy - 2ydy - dy = 2xdx - ydx + dx$

Solve the differential equation with integrating factor. Given the solution passes through $(\pi/2, 8)$, find the minimum value of $y$ where $y = \dfrac{8(1-\cos x)}{\sin^2 x}$.

A and B are two separate reservoirs of water. The capacity of A is double that of B. Both the reservoirs are filled completely with water. Water is released simultaneously from both the reservoirs. For each of the reservoirs, the rate of flow out at any instant is proportional to the quantity of water left in the reservoir. After one hour, the quantity of water in A is 1.5 times the quantity of water in B. After how many hours from the time of release of water, do both A and B have the same quantity of water?

The solution of differential equation $\dfrac{dy}{dx} = \dfrac{x^2 + y^2 + 1}{2xy}$ satisfying $y(1) = 0$ is given by:

A curve passes through the point (1, 1) and satisfies the differential equation $\frac{dx}{dy} - \frac{1}{y}x = 3y$. Which of the following points lies on the curve?

The solution of {xy (1 + \cos x) - y} dx + x dy = 0 is

Find the solution of the differential equation x\frac{dy}{dx} + y - x + xy\cot x = 0, where x \neq 0

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

A curve passes through $\left(1,\dfrac{\pi}{6}\right)$. Let the slope at each point $(x,y)$ be $\dfrac{y}{x}+\sec\!\left(\dfrac{y}{x}\right)$, $x>0$. The equation of the curve is

$x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy = 0$

$\frac{dy}{dx} = \left[y + \left(\frac{dy}{dx}\right)\right]^{1/4}$

The solution of the differential equation (y + x - \sqrt{xy}(x + y))dx + (y - \sqrt{xy}(x + y) - x)dy = 0, is

The solution of the differential equation \frac{dy}{dx} = \sin(x + y) + \cos(x + y) is

The solution of the differential equation $\dfrac{dy}{dx} + \dfrac{y}{x}\sec x = \dfrac{\tan x}{2y}$, where \(0 \leq x

If f(x) be a positive, continuous and differentiable on the interval (a, b). If \lim_{x \to a^+} f(x) = 1 and \lim_{x \to b^-} f(x) = 31/4. Also f'(x) = f^3(x) + \frac{1}{f(x)}, then

Curve through \$(1,-2)\$, slope \$=\\dfrac{x^2-2y}{x}\$. Find \$y\$ at \$x=-1\$.

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

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:

\$y=ax^n+bx^{-n}\$ satisfies \$x^2y''-n^2y=0\$. Given \$y(1)=5\$, \$y'(1)=15\$, find \$n\$.

Number of solutions of \$\\dfrac{dy}{dx}=\\dfrac{y+1}{x-1}\$ through \$(1,0)\$.

The solution of the differential equation $y\,dx - (x+2y^2)dy = 0$ is $x = f(y)$. If $f(-1) = 1$, then $f(1)$ is equal to

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

Differential Equations Questions (544)