Question
The solution of differential equation $$yy' = x\left( {\frac{{{y^2}}}{{{x^2}}} + \frac{{f\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}{{f'\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}} \right)$$ is :
A.
$$f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = c{x^2}$$
B.
$${x^2}f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = {c^2}{y^2}$$
C.
$${x^2}f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = c$$
D.
$$f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = \frac{{cy}}{x}$$
Answer :
$$f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = c{x^2}$$
Solution :
The given equation can be written as $$\frac{y}{x}\frac{{dy}}{{dx}} = \left\{ {\frac{{{y^2}}}{{{x^2}}} + \frac{{f\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}{{f'\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}} \right\}$$
The above equation is a homogeneous equation.
Putting $$y = vx,$$ we get
$$\eqalign{
& v\left[ {v + x\frac{{dv}}{{dx}}} \right] = {v^2} + \frac{{f\left( {{v^2}} \right)}}{{f'\left( {{v^2}} \right)}} \cr
& {\text{or }}vx\frac{{dv}}{{dx}} = \frac{{f\left( {{v^2}} \right)}}{{f'\left( {{v^2}} \right)}}\,\,\,\,\left( {{\text{variable separable}}} \right) \cr
& {\text{or }}\frac{{2vf'\left( {{v^2}} \right)}}{{f\left( {{v^2}} \right)}}dv = 2\frac{{dx}}{x} \cr} $$
Now, integrating both sides, we get
$$\eqalign{
& \log \,f\left( {{v^2}} \right) = \log \,{x^2} + \log \,c\,\,\,\left[ {\log \,c = {\text{ constant}}} \right] \cr
& {\text{or }}\log \,f\left( {{v^2}} \right) = \log \,c\,{x^2}\, \cr
& {\text{or }}\,f\left( {{v^2}} \right) = c{x^2} \cr
& {\text{or }}f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = c{x^2} \cr} $$