A curve passes through the point $(1, \pi / 6)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$. Then, the equation of the curve is

$\operatorname{cosec}\left(\frac{y}{x}\right)=\log x+2$

$\sec \left(\frac{2 y}{x}\right)=\log x+2$

$\cos \left(\frac{2 y}{x}\right)=\log x+\frac{1}{2}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

A curve passes through the point

(1, \pi / 6)

. Let the slope of the curve at each point

(x, y)

\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0

. Then, the equation of the curve is

View answer and explanation

Correct option: A

\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}

Option A,

\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}

, is correct because it is the solution to the given differential equation that passes through the specified point.

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)