Question
The solution of primitive integral equation $$\left( {{x^2} + {y^2}} \right)dy = xydx$$ is $$y = y\left( x \right).$$ If $$y\left( 1 \right) = 1$$ and $$\left( {{x_0}} \right) = e,$$ then $${{x_0}}$$ is equal to-
A.
$$\sqrt {2\left( {{e^2} - 1} \right)} $$
B.
$$\sqrt {2\left( {{e^2} + 1} \right)} $$
C.
$$\sqrt 3 \,e$$
D.
$$\sqrt {\frac{{{e^2} + 1}}{2}} $$
Answer :
$$\sqrt 3 \,e$$
Solution :
The given D.E. is $$\left( {{x^2} + {y^2}} \right)dy = xy\,dx\,\,\,{\text{s}}{\text{.t}}{\text{.}}\,\,y\left( 1 \right) = 1$$ and $$y\left( {{x_0}} \right) = e$$
The given equation can be written as
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr
& {\text{Put }}y = vx, \cr
& \therefore v + x\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}} \cr
& \Rightarrow x\frac{{dv}}{{dx}} = \frac{{ - {v^3}}}{{1 + {v^2}}} \cr
& \Rightarrow \int {\frac{{1 + {v^2}}}{{{v^3}}}dv + \int {\frac{{dx}}{x} = 0} } \cr
& \Rightarrow - \frac{1}{{2{v^2}}} + \log \left| v \right| + \log \left| x \right| = C \cr
& \Rightarrow \log \,y = C + \frac{{{x^2}}}{{2{y^2}}}\,\,\,\,\,\,\,\left( {{\text{using }}v = \frac{y}{x}} \right) \cr
& {\text{Also, }}y\left( 1 \right) = 1 \Rightarrow \log 1 = C + \frac{1}{2} \Rightarrow C = - \frac{1}{2} \cr
& \therefore \,\log \,y = \frac{{{x^2} - {y^2}}}{{2{y^2}}},\,\,\,\,\,\,\,\,{\text{But given }}y\left( {{x_0}} \right) = e \cr
& \Rightarrow \log \,e = \frac{{x_0^2 - {e^2}}}{{2{e^2}}}\,\,\,\, \Rightarrow x_0^2 = 3{e^2}\,\,\,\, \Rightarrow {x_0} = \sqrt 3 \,e \cr} $$