MCQ
વિકલ સમીકરણ ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$ નો ઉકેલ મેળવો.
- A$y = 2$
- B$y = 2x$
- ✓$y = 2x - 4$
- D$y = 2{x^2} - 4$
If $\frac{{dy}}{{dx}} = p,$then $y = px - {p^2}$
Differentiating w.r.t. $x$, we get
$p = p + x\frac{{dp}}{{dx}} - 2p\frac{{dp}}{{dx}}$ ==> $\frac{{dp}}{{dx}}(x - 2p) = 0$ ==> $\frac{{dp}}{{dx}} = 0$
Integrating w.r.t. $x$, we get $p = c$
$\therefore \frac{{dy}}{{dx}} = c$; $\therefore y = cx - {c^2}$
If $c = 2$, then $y = 2x - 4$.
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$f\left( x \right) = \int_1^x {\left\{ {2\left( {t - 1} \right){{\left( {t - 2} \right)}^3} + 3{{\left( {t - 1} \right)}^2}{{\left( {t - 2} \right)}^2}} \right\}} dt$ એ $x$ ની કઇ કિમત આગળ મહત્તમ થાય ?