Differential Equations Questions (544)

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $$\int_0^x \bigl(t\cdot f(x) - x\cdot f(t)\bigr)\,t\,dt = 0,\quad \forall\, x\in\mathbb{R}$$ and $f(2) = -\dfrac{1}{3}$. Then:

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $$\int_0^x \bigl(t\cdot f(x) - x\cdot f(t)\bigr)\,t\,dt = 0,\quad \forall\, x\in\mathbb{R}$$ and $f(2) = -\dfrac{1}{3}$. Then:

Let $f(x)$ be continuously differentiable on $(0,\infty)$ with $f(1)=2$ and $\displaystyle\lim_{t\to x}\frac{10\cdot t\cdot f(x) - t^{10}\cdot f(t)}{t-x} = 1$ for all $x>0$. Then $f(x)$ is:

Let $f(x)$ be continuously differentiable on $(0,\infty)$ with $f(1)=2$ and $\displaystyle\lim_{t\to x}\frac{10\cdot t\cdot f(x) - t^{10}\cdot f(t)}{t-x} = 1$ for all $x>0$. Then $f(x)$ is:

If $f(x), g(x)$ be twice differentiable functions on [0, 2] satisfying $f''(x) = g''(x)$, $f'(1) = 2g'(1) = 4$ and $f(2) = 3g(2) = 9$, then:

Through any point $(x, y)$ of a curve which passes through the origin, lines are drawn parallel to the co-ordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of:

If $|y| = f(x)$ is solution of $\frac{d^2y}{dx^2} = \frac{x^3}{y^3}\frac{d^2x}{dy^2}$ such that $f(0) = 2$ and $y = g(x)$ is solution of $\frac{d^2y}{dx^2} + \frac{8y^3}{x^3} + \frac{d^2x}{dy^2} = 0$ such that $g(1) = \frac{1}{3}$, then:

If $\frac{dy}{dx} = \frac{x^2-y}{x+y}$ such that $y = f(x)$ is a solution of differential equation & $f(0) = 0$, then:

Let $y = f(x)$ be a curve in the first quadrant such that the triangle formed by the co-ordinate axis and the tangent at any point on the curve has area 2. If $f(1) = 1$, then $y(2) = $

Let $\frac{dy}{dx}+y = f(x)$ where $y$ is a continuous function of $x$ with $y(0) = 1$ and $f(x) = \begin{cases} e^{-x}, & \text{if } x \le 2 \\ e^{-2}, & \text{if } x > 2 \end{cases}$. Which is of the following hold(s) good?

Solution of the differential equation $x\cos\left(\frac{x}{y}\right)(ydx + xdy) = y\sin\left(\frac{x}{y}\right)(xdy - ydx)$ is :

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

The curve $y = f(x)$ is such that the area of the trapezium formed by the coordinate axes ordinate of an arbitrary point and the tangent at this point equals half the square of its abscissa. The curve is:

Solution of the differential equation $x = 1 + xy\frac{dy}{dx} + \frac{x^2y^2}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{x^3y^3}{3!}\left(\frac{dy}{dx}\right)^3 + ......$ is:

Solution of the differential equation $\left\{1 - \frac{y^2}{x(x-y)^2}\right\}dx + \left\{\frac{x^2}{(x-y)^2} - \frac{1}{y}\right\}dy = 0$ is :

Let $C$ be a curve such that the normal at any point $P$ on it meets $x-$axis $y-$axis at $A$ and $Y$ respectively. If $BP : PA = 1:2$ (internally) and the curve passes through the point $(0,4)$ then which of the following alternative(s) is/are correct?

The solution of $x^3\frac{dy}{dx} + 4x^2\tan y = e^x\sec y$ satisfying $y(1) = 0$ is:

If $y_1, y_2$ are the solution of the differential equation $\frac{dy}{dx} + P(x)y = Q(x)$, then:

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

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

If $f(x) = \int_1^x \frac{\log t}{1 + t + t^2} dt, x \geq 1$, then:

Let $y=y(x)$ be the solution of the differential equation $x^4\,dy+\left(4x^3y+2\sin x\right)dx=0$, $x>0$, $y\!\left(\dfrac{\pi}{2}\right)=0$. Then $\pi^4 y\!\left(\dfrac{\pi}{3}\right)$ is equal to:

Solution of the equation $\frac{xdx + ydy}{xdy - ydx} = \sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ is:

If the length of perpendicular from origin to any normal to the curve $y = f(x)$ is equal to its $y$ intercept, then

If the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of air is 30°C and the substance cools from 37°C to 34°C in 15 min then:

The solution of $y = 2x\left(\frac{dy}{dx}\right) + x^2\left(\frac{dy}{dx}\right)^4$ is :

If differential equation of the curve $y = ae^{3x} + be^{2x} + ce^x$ is $\frac{d^3y}{dx^3} + m\frac{d^2y}{dx^2} + n\frac{dy}{dx} + p y = 0$, then:

Solution of the differential equation $x^2\left(y - x\frac{dy}{dx}\right) = y\left(\frac{dy}{dx}\right)^2$ which does not contain singular solution is :

The orthogonal trajectories of the family of coaxial circles $x^2 + y^2 + 2gx + C = 0$, where $g$ is a parameter are

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

Let $y=y(x)$ be the solution of $\dfrac{dy}{dx}+\dfrac{5}{x(x^5+1)}y=\dfrac{(x^5+1)^2}{x^7}$, $x>0$, with $y(1)=2$. Then $y(2)$ is equal to

The solution of $x^3\frac{dy}{dx} + 4x^2\tan y = e^x\sec y$ satisfying $y(1) = 0$ is:

The orthogonal trajectories of the family of coaxial circles $x^2 + y^2 + 2gx + C = 0$, where $g$ is a parameter are

Let the curve $y = f(x)$ satisfies the equation $\frac{dy}{dx} = 1 - \frac{1}{x}$ and passes through the point $\left(2, \frac{1}{2}\right)$, then the value of $f(1)$ is

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

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

Let $x=x(y)$ be the solution of $2(y+2)\ln_e(y+2)\,dx+(x+4-2\ln_e(y+2))\,dy=0$, $y>-1$ with $x(e^4-2)=1$. Then $x(e^9-2)$ is equal to

Through any point $(x, y)$ of a curve which passes through the origin, lines are drawn parallel to the co-ordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of:

Let $y=y(x)$ be solution of $\dfrac{dy}{dx}+2y=f(x)$, where $f(x)=\begin{cases}1 & x\in[0,1]\\0 & \text{otherwise}\end{cases}$. If $y(0)=0$, then $y(\ln 2)$ is

If $y = f(x)$ is solution of differential equation $\frac{dy}{x dx} - \frac{y}{x^2} = \frac{\sin^{-1}x}{2x^2}$, if $f(0) = 0$, then, $\frac{\pi}{f(1)}$ is equal to ____.

Let $\frac{dy}{dx}+y = f(x)$ where $y$ is a continuous function of $x$ with $y(0) = 1$ and $f(x) = \begin{cases} e^{-x}, & \text{if } x \le 2 \\ e^{-2}, & \text{if } x > 2 \end{cases}$. Which is of the following hold(s) good?

Solution of the differential equation $\left\{1 - \frac{y^2}{x(x-y)^2}\right\}dx + \left\{\frac{x^2}{(x-y)^2} - \frac{1}{y}\right\}dy = 0$ is :

Let $y=y(x)$ be the solution of the differential equation $\sec x\dfrac{dy}{dx}-2y=2+3\sin x$, $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, $y(0)=-\dfrac{7}{4}$. Then $y\!\left(\dfrac{\pi}{6}\right)$ is equal to:

Tangent to a curve intersects the $y$-axis at point $P$. A line perpendicular to this tangent through $P$ passes through the point $(1, 0)$. The differential equation of the curve is

The degree of the differential equation satisfied by the curves $\sqrt{1 + x} - a\sqrt{1 + y} = 1$, is ____.

Let $y=y(x)$ be a differentiable function in the interval $(0,\infty)$ such that $y(1)=2$, and $\displaystyle\lim_{t\to x}\left(\frac{t^2 y(x)-x^2 y(t)}{x-t}\right)=3$ for each $x>0$. Then $2y(2)$ is equal to

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

The equation of the curve passing through the point $(1,1)$ and satisfying the differential equation $\frac{dy}{dx} = \frac{x + 2y - 1}{2x + 1}$ is

The general solution of the differential equation $x\left(\frac{dy}{dx}\right) = y\log\left(\frac{y}{x}\right)$ is:

Solution of the differential equation $x^2\left(y - x\frac{dy}{dx}\right) = y\left(\frac{dy}{dx}\right)^2$ which does not contain singular solution is :