Question
The equation of the curve satisfying $$x\,dy - y\,dx = \sqrt {{x^2} - {y^2}} $$ and $$y\left( 1 \right) = 0$$ is :
A.
$$y = {x^2}\log \left( {\sin \,x} \right)$$
B.
$$y = x\,\sin \left( {\log \,x} \right)$$
C.
$${y^2} = x{\left( {x - 1} \right)^2}$$
D.
$$y = 2{x^2}\left( {x - 1} \right)$$
Answer :
$$y = x\,\sin \left( {\log \,x} \right)$$
Solution :
The given equation can be rewritten as
$$\eqalign{
& {x^2}\left[ {\frac{{x\,dy - y\,dx}}{{{x^2}}}} \right] = x\sqrt {1 - \frac{{{y^2}}}{{{x^2}}}} \cr
& \Rightarrow {x^2}\frac{d}{{dx}}\left( {\frac{y}{x}} \right) = x\sqrt {1 - \frac{{{y^2}}}{{{x^2}}}} \cr
& {\text{Put }}\frac{y}{x} = z,{\text{ we get}} \cr
& \Rightarrow {x^2}\frac{{dz}}{{dx}} = x\sqrt {1 - {z^2}} \cr
& \Rightarrow \frac{{dz}}{{\sqrt {1 - {z^2}} }} = \frac{{dx}}{x} \cr
& {\text{Integrating, we get}} \cr
& {\sin ^{ - 1}}\left( z \right) = \log \,x + c \cr
& \Rightarrow {\sin ^{ - 1}}\left( {\frac{y}{x}} \right) = \log \,x + c \cr
& {\text{Apply the boundary value}}\,y\left( 1 \right) = 0 \cr
& \Rightarrow {\sin ^{ - 1}}\left( {\frac{0}{1}} \right) = \log \,1 + c \Rightarrow c = 0 \cr
& \therefore \,{\sin ^{ - 1}}\left( {\frac{y}{x}} \right) = \log \,x + 0 \Rightarrow y = x\,\sin \left( {\log \,x} \right) \cr} $$