Question

Find the equation of a curve passing through $(1,\frac{\pi }{4})$ if the slope of the tangent to the curve at any point $p(x,y)$ is $\frac{y}{x}-\cos {}^2\frac{y}{x}$.

Answer 1

According to the given condition
slope of tangent, $m=dy/dx$
$\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\frac{y}{x}....\left(i\right)$
This is a homogeneous differential equation. Substituting $y=vx$, we get
$v+x\frac{dv}{dx}=v-{\cos }^{2}v\Rightarrow x\frac{dv}{dx}=-{\cos }^{2}v$
$\Rightarrow {\sec }^{2}vdv=-\frac{dx}{x}\Rightarrow \tan v=-\log x+c\int {\sec }^{2}vdv=\int -\frac{dx}{x}$
$\Rightarrow \tan \frac{y}{x}+\log x=c...\left(ii\right)$
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 equation.