The slope of tangent at any point $(x, y)$ on a curve $y=y(x)$ is $\frac{x^2+y^2}{2 x y}, x>0$. If $y(2)=0$, then a value of $y(8)$ is

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The slope of tangent at any point

(x, y)

on a curve

y=y(x)

\frac{x^2+y^2}{2 x y}, x>0

. If

y(2)=0

, then a value of

y(8)

View answer and explanation

Correct option: D

4 \sqrt{3}

Option D (

4 \sqrt{3}

) is correct because it is the solution to the differential equation given the initial condition y(2) = 0.

Practice All Chapter MCQs →

y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

{x^2} + {y^2} = 1

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

y=y_{1}(x)

y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)