Question
The solution of primitive integral equation $$\left( {{x^2} + {y^2}} \right)dy = xy.dx$$ is $$y = y\left( x \right).$$ If $$y\left( 1 \right) = 1$$ and $$y\left( {{x_0}} \right) = e$$ then $${{x_0}}$$ is :
A.
$$\sqrt {2\left( {{e^2} - 1} \right)} $$
B.
$$\sqrt {2\left( {{e^2} + 1} \right)} $$
C.
$$\sqrt 3 e$$
D.
$$\sqrt {\frac{1}{2}\left( {{e^2} + 1} \right)} $$
Answer :
$$\sqrt 3 e$$
Solution :
$$\eqalign{
& {\text{We have, }}\left( {{x^2} + {y^2}} \right)dy = xy\,dx \cr
& \Rightarrow \frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr
& {\text{Substitute }}y = vx \Rightarrow \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}} \cr
& \Rightarrow v + x\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}} \cr
& \Rightarrow x\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}} - v = - \frac{{{v^3}}}{{1 + {v^2}}} \cr
& \Rightarrow \frac{{1 + {v^2}}}{{{v^3}}}dv = - \frac{{dx}}{x} \cr} $$
Integrating both sides we get, $$ - \frac{1}{{2{v^2}}} + \log \,v = - \log \,x + \log \,c,$$ where $$c$$ is constant
$$\eqalign{
& \Rightarrow - \frac{1}{{2{v^2}}} + \log \left( {\frac{{vx}}{c}} \right) = 0 \cr
& \Rightarrow y = c{e^{\frac{{{x^2}}}{{2{y^2}}}}} \cr
& {\text{Now using }}y\left( 1 \right) = 1 \Rightarrow c = {e^{ - \frac{1}{2}}} \cr
& {\text{Also, }}y\left( {{x_2}} \right) = e \Rightarrow e = {e^{ - \frac{1}{2} + \frac{{x_0^2}}{{2{e^2}}}}} \cr
& \Rightarrow 1 = - \frac{1}{2} + \frac{{x_0^2}}{{2{e^2}}} \cr
& \Rightarrow x_0^2 = 3{e^2} \cr
& \Rightarrow {x_0} = \sqrt 3 e \cr} $$