MCQ
If $y = \frac{x}{{\ln \,|c\,x|}}$ (where $c$ is an arbitrary constant) is the general solution of the differential equation $\frac{{dy}}{{dx}} =  \frac{y}{x}+  \phi \left( {\frac{x}{y}} \right)$ then the function $\phi \left( {\frac{x}{y}} \right)$ is :
  • A
    $\frac{{{x^2}}}{{{y^2}}}$
  • B
    $- \frac{{{x^2}}}{{{y^2}}}$
  • C
    $\frac{{{y^2}}}{{{x^2}}}$
  • $ - \frac{{{y^2}}}{{{x^2}}}$

Answer

Correct option: D.
$ - \frac{{{y^2}}}{{{x^2}}}$
d
$\ln c + \ln |x| = \frac{x}{y}$
diff. $w.r.t.\,\, x,\,\, \frac{1}{x} = \frac{{y - x\,{y_1}}}{{{y^2}}}$
$\frac{{{y^2}}}{x} = y - x\frac{{dy}}{{dx}}$
$\frac{{dy}}{{dx}} = \frac{y}{x} - \frac{{{y^2}}}{{{x^2}}}$
$\Rightarrow \phi \left( {\frac{x}{y}} \right) = - \frac{{{y^2}}}{{{x^2}}}$

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