JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Medium

1 of 1

Application +4 | -1

A curve passes through the point

\left(1, \frac{\pi}{4}\right)

and its slope at any point is given by

\frac{y}{x}-\cos ^2\left(\frac{y}{x}\right)

. Then, the curve has the equation

View answer and explanation

Correct option: A

y=x \tan ^{-1}\left(\ln \frac{e}{x}\right)

Option A,

y=x \tan ^{-1}(\ln \frac{e}{x})

, is the correct equation of the curve that satisfies the given differential equation and passes through the point

(1, \frac{\pi}{4})

More Questions from DIFFERENTIAL EQUATIONS

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)