If x=f(y) is the solution of the differential equation (1+y^2)+(x-2e^ ^-1y)(dy)/(dx)=0,y (…

Question

If $x=f(y)$ is the solution of the differential equation $(1+{y}^{2})+(x-2{e}^{{	an }^{-1}y})\frac{dy}{dx}=0,y\in (-\frac{\pi }{2},\frac{\pi }{2})$ with $f(0)=1$, then $f(\frac{1}{\sqrt{3}})$ is equal to :

Accepted Answer

$\begin{gathered} \frac{dx}{dy}+\frac{x}{1+{y}^{2}}=\frac{2{e}^{{	an }^{-1}y}}{1+{y}^{2}} \ ={e}^{{	an }^{-1}y} \ {x}^{{	an }^{-1}y}=\int \frac{2{({e}^{{	an }^{-1}y})}^{2}dy}{1+{y}^{2}} \ {	an }^{-1}y=t,\frac{dy}{1+{y}^{2}}=dt \ x{e}^{{	an }^{-1}y}=\int 2{e}^{2t}dt \ x{e}^{{	an }^{-1}y}={e}^{2{	an }^{-1}y}+c \ x={e}^{{	an }^{-1}y}+c{e}^{-{	an }^{-1}y} \ \because y=0,x=1 \ 1=1+c\Rightarrow c=0 \ y=\frac{1}{\sqrt{3}},x={e}^{\pi /6} \end{gathered}$

Solutions · 1