question_answer

Find the equation of a curve passing through \[\left( 1,\,\,\frac{\pi }{4} \right),\] if the slop of the tangent to the curve at any point \[P\left( x,\text{ }y \right)\]is \[\frac{y}{x}-{{\cos }^{2}}\frac{y}{x}.\]

Answer:

According to the given condition, we have \[\frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}\frac{y}{x}=f\left( \frac{y}{x} \right),\] Which is a homogeneous differential equation. Now, putting \[y=vx\] and \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] in the above equation, we get \[v+x\frac{dv}{dx}=v-{{\cos }^{2}}v\] \[\Rightarrow \] \[x\frac{dv}{dx}=-{{\cos }^{2}}v\] \[\Rightarrow \]   \[{{\sec }^{2}}v\,dv=\frac{-dx}{x}\] On integrating both sides, we get \[\tan \,v=-\log |x|+C\] \[\Rightarrow \] \[\tan v+\log |x|=C\] \[\Rightarrow \,\,\,\tan \left( \frac{y}{x} \right)+\log |x|=C\]                       \[\left[ \text{put}\,v=\frac{y}{x} \right]\] Since, the curve passes through \[\left( 1,\,\,\frac{\pi }{4} \right),\] therefore we have, \[\tan \,\,\left( \frac{\pi }{4} \right)+\log \,|1|\,\,=C\] \[\Rightarrow \]   C = 1       \[\left[ \because \,\,\,\log 1=0\,\,and\,\,tan\frac{\pi }{4}=1 \right]\] Hence, the required equation of curve is \[\tan \left( \frac{y}{x} \right)+\log |x|\,\,=1\]