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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{\phi \,\left( {\frac{y}{x}} \right)}}{{\phi '\,\left( {\frac{y}{x}} \right)}}$ is
  • ✓
    $\phi \,\left( {\frac{y}{x}} \right) = kx$
  • B
    $x\,\phi \,\left( {\frac{y}{x}} \right) = k$
  • C
    $\phi \,\left( {\frac{y}{x}} \right) = ky$
  • D
    $y\,\phi \left( {\frac{y}{x}} \right) = k$
✓

Answer

Correct option: A.
$\phi \,\left( {\frac{y}{x}} \right) = kx$
a
(a) $\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{\phi \,\left( {\frac{y}{x}} \right)}}{{\phi '\,\left( {\frac{y}{x}} \right)}}$. Put $y = vx$ ==> $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

 The given differential equation becomes

$v + x\frac{{dv}}{{dx}} = v + \frac{{\phi \,(v)}}{{\phi '\,(v)}}$ ==> $\frac{{\phi '(v)}}{{\phi (v)}}dv = \frac{{dx}}{x}$

==> $\log \phi (v) = \log x + \log k$ ==> $\phi (v) = kx$ ==> $\phi \,\left( {\frac{y}{x}} \right) = kx$.

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