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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
Solution of the equation $(x + \log y)dy + y\,dx = 0$ is
  • A
    $xy + y\log y = c$
  • ✓
    $xy + y\log y - y = c$
  • C
    $xy + \log y - x = c$
  • D
    None of these
✓

Answer

Correct option: B.
$xy + y\log y - y = c$
b
(b) $xdy + ydx + \log ydy = 0$ ==> $xdy + ydx = - \log ydy$

$y\frac{{dx}}{{dy}} + x = - \log y$ ==> $\frac{{dx}}{{dy}} + \frac{x}{y} = - \frac{{\log y}}{y}$

$I.F.$ =${e^{\int {\frac{1}{y}dy} }} = y$

Hence solution is $x.y = - \int {y.\frac{{\log ydy}}{y} + c} $

==> $xy = - (y\log y - y) + c$ ==> $xy + (y\log y - y) = c$.
 

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