Question
If $$x\frac{{dy}}{{dx}} + y = x.\frac{{f\left( {x.y} \right)}}{{f'\left( {x.y} \right)}}$$ then $$f\left( {x.y} \right)$$ is equal to ($$k$$ being an arbitrary constant) :
A.
$$k{e^{\frac{{{x^2}}}{2}}}$$
B.
$$k{e^{\frac{{{y^2}}}{2}}}$$
C.
$$k{e^{\frac{{xy}}{2}}}$$
D.
none of these
Answer :
$$k{e^{\frac{{{x^2}}}{2}}}$$
Solution :
$$\eqalign{
& d\left( {xy} \right) = \frac{{f\left( {xy} \right)}}{{f'\left( {xy} \right)}}x\,dx \cr
& {\text{or }}\frac{{f'\left( {xy} \right)}}{{f\left( {xy} \right)}}d\left( {xy} \right) = x\,dx\,\,\, \Rightarrow \int {\frac{{f'\left( {xy} \right)}}{{f\left( {xy} \right)}}d\left( {xy} \right)} = \int {x\,dx} \cr
& {\text{or }}\log \left\{ {f\left( {xy} \right)} \right\} = \frac{{{x^2}}}{2} + c \cr
& \therefore f\left( {xy} \right) = {e^{\frac{{{x^2}}}{2} + c}} = {e^c}.{e^{\frac{{{x^2}}}{2}}} = k{e^{\frac{{{x^2}}}{2}}} \cr} $$