MCQ
$({x^2} + {y^2})dy = xydx$. જો $y({x_0}) = e$, $y(1) = 1$, તો ${x_0} = $
- ✓$\sqrt 3 e$
- B$\sqrt {{e^2} - \frac{1}{2}} $
- C$\sqrt {\frac{{{e^2} - 1}}{2}} $
- D$\sqrt {\frac{{{e^2} + 1}}{2}} $
==> $x\frac{{(ydx - xdy)}}{{{y^2}}} = dy$ ==> $\frac{x}{y}d\left( {\frac{x}{y}} \right) = \frac{{dy}}{y}$
Integrating, $\frac{{{x^2}}}{{2{y^2}}} = {\log _e}y + c$
Given $y(1) = 1$ ==> $c = \frac{1}{2}$ ==> $\frac{{{x^2}}}{{2{y^2}}} = {\log _e}y + \frac{1}{2}$
Now $y({x_0}) = e$ ==> $\frac{{x_0^2}}{{2{e^2}}} - {\log _e}e - \frac{1}{2} = 0$ ==> $x_0^2 = 3{e^2}$
==> ${x_0} = \pm \sqrt 3 e$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.