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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
The solution of the differential equation $x + y\frac{{dy}}{{dx}} = 2y$ is
  • A
    $\log (y - x) = c + \frac{{y - x}}{x}$
  • ✓
    $\log (y - x) = c + \frac{x}{{y - x}}$
  • C
    $y - x = c + \log \frac{x}{{y - x}}$
  • D
    $y - x = c + \frac{x}{{y - x}}$
✓

Answer

Correct option: B.
$\log (y - x) = c + \frac{x}{{y - x}}$
b
(b) Given $x + y\frac{{dy}}{{dx}} = 2y$ ==> $\frac{x}{y} + \frac{{dy}}{{dx}} = 2$

Put $y = vx$and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

$\therefore \frac{1}{v} + v + x\frac{{dv}}{{dx}} = 2$ ==> $v + x.\frac{{dv}}{{dx}} = \frac{{2v - 1}}{v}$

==> $\frac{v}{{{{(v - 1)}^2}}}dv = - \frac{{dx}}{x}$ ==> $\frac{{v - 1 + 1}}{{{{(v - 1)}^2}}}dv = - \frac{{dx}}{x}$

$\left[ {\frac{1}{{(v - 1)}} + \frac{1}{{{{(v - 1)}^2}}}} \right]dv = - \frac{{dx}}{x}$

Integrating both sides, $\int_{}^{} {\frac{{dv}}{{v - 1}}} + \int_{}^{} {\frac{{dv}}{{{{(v - 1)}^2}}}} = - \int_{}^{} {\frac{{dx}}{x}} $

==> $\log (v - 1) - \frac{1}{{v - 1}} = - \log x + c$ ==> $\log (y - x) = \frac{x}{{y - x}} + c$.

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