Question

A curve passes through the point $\left( {1,\dfrac{\pi }{4}} \right)$ and its slope at any point is given by $\dfrac{y}{x} - {\cos ^2}\left( {\dfrac{y}{x}} \right)$ . Then the curve is
A) $y = x{\tan ^{ - 1}}\ln \dfrac{e}{x}$
B) $y = x{\tan ^{ - 1}}(\ln + 2)$
C) $y = \dfrac{1}{x}{\tan ^{ - 1}}\ln \dfrac{e}{x}$
D) None

Answer 1

As we know that the Slope of any curve is $\dfrac{{dy}}{{dx}}$ hence from this we equate it as $\dfrac{{dy}}{{dx}}$ = $\dfrac{y}{x} - {\cos ^2}\left( {\dfrac{y}{x}} \right)$ now let us suppose that the $\dfrac{y}{x} = v$ put it in the above equation and integrate it and get the curve equation .

Complete step-by-step answer:
As we know that the slope of the curve at any arbitrary point $(x,y)$ is $\dfrac{{dy}}{{dx}}$ .
It is given in the question that slope at any point is $\dfrac{y}{x} - {\cos ^2}\left( {\dfrac{y}{x}} \right)$ ,
Hence it is equal to
$\dfrac{{dy}}{{dx}}$ = $\dfrac{y}{x} - {\cos ^2}\left( {\dfrac{y}{x}} \right)$
It is homogeneous differential equation to solve according ,
So now we have to solve this differential equation ,
for this let us take $\dfrac{y}{x} = v$ hence ,
$y = vx$
Differentiate with respect to x
$\dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v\dfrac{{dx}}{{dx}}$
$\dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v$
Now put it in the given differential equation ,
$x\dfrac{{dv}}{{dx}} + v$ = $v - {\cos ^2}\left( v \right)$
So $v$ is common on both side hence it will cancel out remaining equation become ,
$x\dfrac{{dv}}{{dx}}$ = $- {\cos ^2}\left( v \right)$
$\dfrac{{dv}}{{{{\cos }^2}v}} = - \dfrac{1}{x}dx$
${\sec ^2}vdv = - \dfrac{1}{x}dx$
Now integrate on both side ,
$\int {{{\sec }^2}vdv} = - \int {\dfrac{1}{x}dx}$
Integration of ${\sec ^2}v$ is $\tan v$ and Integration of $\dfrac{1}{x}$ is $\ln x$ ,
So ,
$\tan v = - \ln x + C$
As initially we suppose that $\dfrac{y}{x} = v$ on putting the value
$\tan \dfrac{y}{x} = - \ln x + C$
As it is given that the curve is passes to the point $\left( {1,\dfrac{\pi }{4}} \right)$ hence put the value of this is
$\tan \dfrac{\pi }{4} = \ln 1 + C$
we know that the $\tan \dfrac{\pi }{4} = 1$ and $\ln 1 = 0$
$C = 1$
Hence the equation of curve is $\tan \dfrac{y}{x} = - \ln x + 1$
As we can write $1 = \ln e$
$\tan \dfrac{y}{x} = \ln e - \ln x$
From property of log that $\log A - \log B = \log \dfrac{A}{B}$ so use it in above equation we get ,
$\tan \dfrac{y}{x} = \ln \dfrac{e}{x}$
$y = x{\tan ^{ - 1}}\ln \dfrac{e}{x}$

So, the correct answer is “Option A”.

Note: Any curve, which cuts every member of a given family of curves at right angle, is called an orthogonal trajectory of the family.
Procedure for finding the Orthogonal Trajectory
Let $f(x,y,c) = 0$ is the given equation of the family of curves, where ‘c’ is an arbitrary parameter.
Now Differentiate the equation , with respect to x and try to eliminate $0$ , that is from a differential equation.
Now Substitute $- \dfrac{{dx}}{{dy}}$ for $\dfrac{{dy}}{{dx}}$ in the given differential equation.Hence this will be the given differential equation of the orthogonal trajectories on solving we get the required orthogonal trajectories .