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
$v+x\frac{dv}{dx}=v-{\cos }^{2}v$
$\Rightarrow x\frac{dv}{dx}=-{\cos }^{2}v$
$\Rightarrow \int {\sec }^{2}vdv=-\int \frac{dx}{x}$
$\Rightarrow \tan v=-\log x+C$
$\Rightarrow \tan \frac{y}{x}+\log x=C$
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