MCQ
The solution of the equation $\frac{{dy}}{{dx}} = \frac{x}{{2y - x}}$ is
- ✓$(x - y){(x + 2y)^2} = c$
- B$y = x + c$
- C$y = (2y - x) + c$
- D$y = \frac{x}{{2y - x}} + c$
✓
Answer
Correct option: A.
$(x - y){(x + 2y)^2} = c$
a
(a) $\frac{{dy}}{{dx}} = \frac{x}{{2y - x}}$. Put $y = vx$ ==> $v + x\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}}$
(a) $\frac{{dy}}{{dx}} = \frac{x}{{2y - x}}$. Put $y = vx$ ==> $v + x\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}}$
$v + x\frac{{dv}}{{dx}} = \frac{x}{{2v - x}} = \frac{1}{{2v - 1}}$
$x\frac{{dv}}{{dx}} = \frac{1}{{2v - 1}} - v = \frac{{1 - 2{v^2} + v}}{{2v - 1}} = - \frac{{(v - 1)(2v + 1)}}{{2v - 1}}$
$\frac{{(2v - 1)}}{{(2v + 1)(v - 1)}} = \frac{{ - dx}}{x}$; $\frac{1}{{3(v - 1)}} + \frac{4}{{3(2v + 1)}} = \frac{{ - dx}}{x}$
$\frac{1}{3}\log (v - 1) + \frac{4}{3}.\frac{1}{2}\log (2v + 1) = \log \frac{1}{x} + \log c$
$\log {(v - 1)^{1/3}} + \log {(2v + 1)^{2/3}} = \log \frac{c}{x}$
$ = {(v - 1)^{1/3}}{(2v + 1)^{2/3}} = \frac{c}{x}$
$\left( {\frac{{y - x}}{x}} \right){\rm{ }}{\left( {\frac{{2y + x}}{x}} \right)^2} = \frac{{{c^3}}}{{{x^3}}}$ ==> $(x - y){(x + 2y)^2} = c$.
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