The slope of the tangent at $\left( {x,y} \right)$ to a curve passing through a point $\left( {2,1} \right)$ is $\frac{{{x^2} + {y^2}}}{{2xy}}$, then the equation of the curve is

$2\left( {{x^2} - {y^2}} \right) = 6y$

$x\left( {{x^2} - {y^2}} \right) = 6$

$x\left( {{x^2} + {y^2}} \right) = 10$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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1 of 1

Application +4 | -1

The slope of the tangent at

\left( {x,y} \right)

to a curve passing through a point

\left( {2,1} \right)

\frac{{{x^2} + {y^2}}}{{2xy}}

, then the equation of the curve is

View answer and explanation

Correct option: A

2\left( {{x^2} - {y^2}} \right) = 3x

Option A (

2\left( {{x^2} - {y^2}} \right) = 3x

) is correct because it is the equation of the curve obtained by solving the differential equation and applying the given point (2, 1).

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)