Q53 / 125 · BITSAT 2021 · Mathematics
If the slope of the tangent to the curve at any point $\mathrm{P}(x, y)$ is $\frac{y}{x}-\cos ^{2} \frac{y}{x}$, then the equation of a curve passing through $\left(1, \frac{\pi}{4}\right)$ is
- Correct answerA$\tan \left(\frac{y}{x}\right)+\log x=1$
- B$\tan \left(\frac{y}{x}\right)+\log y=1$
- C$\tan \left(\frac{x}{y}\right)+\log x=1$
- D$\tan \left(\frac{x}{y}\right)+\log y=1$
Solution
According to the condition,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{x}-\cos ^{2} \frac{y}{x} \ldots \text { (i) }$
This is a homogeneous differential equation Substituting $y=v x$, we get
$\begin{array}{l}
v+x \frac{\mathrm{dv}}{\mathrm{dx}}=v-\cos ^{2} v \\
\Rightarrow x \frac{\mathrm{dv}}{\mathrm{dx}}=-\cos ^{2} v \\
\Rightarrow \int \sec ^{2} \mathrm{vdv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\
\Rightarrow \tan v=-\log x+\mathrm{C} \\
\Rightarrow \tan \frac{y}{x}+\log x=\mathrm{C}
\end{array}$
Substituting $\mathrm{x}=1, y=\frac{\pi}{4}$, we get $\mathrm{C}=1$
Thus, we get
$\tan \left(\frac{y}{x}\right)+\log x=1$
which is the required solution.
Hence, the correct answer is option: 'A'.