The solution of $(1+xy)y \, dx + (1-xy)x \, dy = 0$ is:
- A$\log \left(\frac{x}{y}\right) + \frac{1}{xy} = k$,where $k$ is the constant of integration.
- B$\log \left(\frac{x}{y}\right) = \frac{1}{xy} + k$,where $k$ is the constant of integration.
- C$\log \left(\frac{x}{y}\right) + xy = k$,where $k$ is the constant of integration.
- D$\log \left(\frac{x}{y}\right) = xy + k$,where $k$ is the constant of integration.
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Let $\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$,where $a, b, c$ are constants,represent a circle passing through the point $(2, 5)$. Then the shortest distance of the point $(11, 6)$ from this circle is
DifficultJEE MAIN 2022
View SolutionMatch the statements/expressions in Column $I$ with the open intervals in Column $II$.
Column $I$ Column $II$ $(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$ $(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$ $(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$ $(q)$ $(0, \frac{\pi}{2})$ $(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies $(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$ $(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing $(s)$ $(0, \frac{\pi}{8})$ $(t)$ $(-\pi, \pi)$
| Column $I$ | Column $II$ |
| $(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$ | $(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$ |
| $(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$ | $(q)$ $(0, \frac{\pi}{2})$ |
| $(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies | $(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$ |
| $(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing | $(s)$ $(0, \frac{\pi}{8})$ |
| $(t)$ $(-\pi, \pi)$ |
Let the tangent at any point $P(x, y)$ on a curve passing through the points $(1, 1)$ and $(\frac{1}{10}, 100)$ intersect the positive $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If $PA: PB = 1: k$ and $y = y(x)$ is the solution of the differential equation $e^{\frac{dy}{dx}} = 2x + 1$ with $y(0) = 2$,then $4y(1) - 5 \log_e 3$ is equal to:
The differential equation $\frac{dx}{dy} = \frac{3y}{2x}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity:
The solution of the differential equation $y dx - x dy + 3x^2 y^2 e^{x^3} dx = 0$ satisfying $y = 1$ when $x = 1$ is:
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