STD 12 - 9. differential equations — JEE कक्षा 11 साइन्स — Question

$y = vx$ ==> $v + x\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}}$ रखने पर,

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

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

$\frac{{(2v - 1)}}{{(2v + 1)(v - 1)}} = \frac{{ - dx}}{x}$; $\frac{1}{{3(v - 1)}} + \frac{4}{{3(2v + 1)}} = \frac{{ - dx}}{x}$

$\frac{1}{3}\log (v - 1) + \frac{4}{3}.\frac{1}{2}\log (2v + 1) = \log \frac{1}{x} + \log c$

$\log {(v - 1)^{1/3}} + \log {(2v + 1)^{2/3}} = \log \frac{c}{x}$

==> ${(v - 1)^{1/3}}{(2v + 1)^{2/3}} = \frac{c}{x}$

$\left( {\frac{{y - x}}{x}} \right){\rm{ }}{\left( {\frac{{2y + x}}{x}} \right)^2} = \frac{{{c^3}}}{{{x^3}}}$ ==> $(x - y){(x + 2y)^2} = c$