Question
The differential equation $$\phi \left( x \right)dy = y\left\{ {\phi '\left( x \right) - y} \right\}dx$$ is changed in the form $$df\left( {x,\,y} \right) = 0.$$ Then $$f\left( {x,\,y} \right)$$ is :
A.
$$\frac{1}{2}\phi \left( x \right) + y$$
B.
$$\frac{1}{y}\phi \left( x \right) - x$$
C.
$$\frac{1}{y}\phi \left( x \right) + x$$
D.
$$\frac{{\phi \left( x \right)}}{y}$$
Answer :
$$\frac{1}{y}\phi \left( x \right) - x$$
Solution :
$$\eqalign{
& \phi \left( x \right)dy = y\phi \left( x \right)dx - {y^2}dx \cr
& {\text{or }}\frac{{y\phi '\left( x \right)dx - \phi \left( x \right)dy}}{{{y^2}}}dy - dx = 0 \cr
& {\text{or }}d\left( {\frac{{\phi \left( x \right)}}{y}} \right) - dx = 0 \cr
& {\text{or }}d\left( {\frac{{\phi \left( x \right)}}{y} - x} \right) = 0 \cr} $$