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

$\frac{{dx}}{{dy}} + {\left( {\frac{x}{y}} \right)^2} - \left( {\frac{x}{y}} \right) + 1 = 0$

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

$v + y\frac{{dv}}{{dy}} + {v^2} - v + 1 = 0$ ==> $\frac{{dv}}{{{v^2} + 1}} + \frac{{dy}}{y} = 0$

==> $\int {\frac{{dv}}{{{v^2} + 1}} + \int {\frac{{dy}}{y} = 0} } $ ==> ${\tan ^{ - 1}}(v) + \log y + C = 0$

==> ${\tan ^{ - 1}}(x/y) + \log y + c = 0$