Question

If the slope of the tangent to the curve at any point $P ( x , y )$ is $\frac{y}{x}-\cos ^{2} \frac{y}{x}$, then the equation of a curve passing through $\left(1, \frac{\pi}{4}\right)$ is

• A. $\tan \left(\frac{ y }{ x }\right)+\log x =1$
• B. $\tan \left(\frac{ y }{ x }\right)+\log y =1$
• C. $\tan \left(\frac{ x }{ y }\right)+\log x =1$
• D. $\tan \left(\frac{x}{y}\right)+\log y=1$

Answer 1

Answer: (A)

According to the condition,
$\frac{ dy }{ dx }=\frac{ y }{ x }-\cos ^{2} \frac{ y }{ x } \dots$(i)
This is a homogeneous differential equation
Substituting $y = vx$, we get
$v + x \frac{ d v }{ dx }= v -\cos ^{2} v $
$\Rightarrow x \frac{ d v }{ dx }=-\cos ^{2} v$
$\Rightarrow \int \sec ^{2} v d v=-\int \frac{d x}{x} $
$\Rightarrow \tan v=-\log x+C$
$\Rightarrow \tan \frac{y}{x}+\log x=C$
Substituting $x =1, y =\frac{\pi}{4}$,
we get $C =1$. Thus, we get
$\tan \left(\frac{y}{x}\right)+\log x=1$
which is the required solution