y = x x + 1 is a solution of the differential equation - Vidyadip

MCQ

$y = \frac{x}{{x + 1}}$ is a solution of the differential equation

A
${y^2}\frac{{dy}}{{dx}} = {x^2}$
✓
${x^2}\frac{{dy}}{{dx}} = {y^2}$
C
$y\frac{{dy}}{{dx}} = x$
D
$x\frac{{dy}}{{dx}} = y$

✓

Answer

Correct option: B.

${x^2}\frac{{dy}}{{dx}} = {y^2}$

b
(b) $y = \frac{x}{{x + 1}} \Rightarrow \frac{1}{y} = 1 + \frac{1}{x}$

$ - \frac{1}{{{y^2}}}\frac{{dy}}{{dx}} = 0 - \frac{1}{{{x^2}}}$ ==> ${x^2}\frac{{dy}}{{dx}} = {y^2}$.

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

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The larger of ${99^{50}} + {100^{50}}$ and ${101^{50}}$ is

If $\theta$ is eliminated from the equations $x = a \,cos(\theta - \alpha )$ and $y = b\, cos (\theta - \beta )$ then $\frac{{{x^2}}}{{{a^2}}}\,\, + \,\,\frac{{{y^2}}}{{{b^2}}}\,\, - \,\,\frac{{2xy}}{{ab}}\,\cos (\alpha - \beta )\,$ is equal to

The set of all $\alpha$, for which the vector $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k} \quad$ and $\quad \vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k} \quad$ are inclined at an obtuse angle for all $t \in \mathbb{R}$ is :

If the projections of a line segment on the $x, y$ and $z-$ axes in $3-$ dimensional space are $2, 3$ and $6$ respectively, then the length of the line segment is

$16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)$ is equal to

A line parallel to the straight line $2 x-y=0$ is tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point $\left(x_{1}, y_{1}\right) .$ Then $x_{1}^{2}+5 y_{1}^{2}$ is equal to

If $y = a + b{x^2}$; $a$ , $b$ arbitrary constants, then

Let ${I_n} = \smallint {\tan ^n}xdx,\left( {n > 1} \right).$ ${I_4} + {I_6} = a{\tan ^5}x + b{x^5} + C$, where $C$ is constant of integration , then ordered pair $\left( {a,b} \right)$ equal to :

The equation of the circle of radius $5$ and touching the coordinate axes in third quadrant is

$\int_{}^{} {\frac{{{x^4}}}{{(x - 1)({x^2} + 1)}}dx = } $