MCQ
The solution of the differential equation $\frac{dy}{dx}= \frac{y}{(y^2-x)}$ is
  • A
    $y^3 -xy = c$
  • $y^3 -3xy = c$
  • C
    $y^3 + 3xy = c$
  • D
    $y^3 + xy = c$

Answer

Correct option: B.
$y^3 -3xy = c$
b
$\frac{d x}{d y}=y-\frac{x}{y} \Rightarrow \frac{d x}{d y}+\frac{1}{y} \cdot x=y$

$\therefore $ solution is $x . y=\int y^{2} d y+c \Rightarrow x . y=\frac{y^{3}}{3}+c$

$\Rightarrow \mathrm{y}^{3}-3 \mathrm{xy}=\mathrm{c}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$ is equal to:
  1. $0$
  2. ${\pi}$
  3. $\frac{\pi}{2}$
  4. $\frac{\pi}{4}$
The area bounded by the curve $y=\cos x$, the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ is
Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals
If the distance  $‘s’ $ metre traversed by a particle in $ t$  seconds is given by $s = {t^3} - 3{t^2}$, then the velocity of the particle when the acceleration is zero, in $metre/sec$ is
If $y = {x^2} + {1 \over {{x^2} + {1 \over {{x^2} + {1 \over {{x^2} + ......\infty }}}}}},$ then ${{dy} \over {dx}} = $
For the $LP$ problem

Maximize $z=2 x+3 y$ the coordinates of the corner points of the bounded feasible region are $A\,(3,3), B\,(20,3),$ $\mathrm{C}\,(20,10), \mathrm{D}\,(18,12)$ and $\mathrm{E}\,(12,12) .$ The maximum value of $z$ is $\ldots \ldots$

Let the area of the region $\{(x, y): x-2 y+4 \geq 0$, $\left.x+2 y^2 \geq 0, x+4 y^2 \leq 8, y \geq 0\right\}$ be $\frac{m}{n}$, where $m$ and $\mathrm{n}$ are coprime numbers. Then $\mathrm{m}+\mathrm{n}$ is equal to
Find the value of $\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$.
Let $f:R \to R$ be a function defined by $f(x) = \frac{{x - m}}{{x - n}}$, where $m \ne n$. Then
The area of the region bounded by y = (x - 4)2, y = 16 - x2 and the x axis, is: