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$y - x = c + \frac{x}{{y - x}}$
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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($A$) There is exactly one choice for such $\vec{v}$
($B$) There are infinitely many choices for such $\vec{v}$
($C$) If $\hat{u}$ lies in the $x y$-plane then $\left|u_1\right|=\left|u_2\right|$
($D$) If $\hat{u}$ lies in the $x z$-plane then $2\left|u_1\right|=\left|u_3\right|$