MCQ
સમીકરણ $\frac{{dy}}{{dx}} = {(x + y)^2}$ નો ઉકેલ મેળવો.
- A$x + y + \tan (x + c) = 0$
- B$x - y + \tan (x + c) = 0$
- ✓$x + y - \tan (x + c) = 0$
- Dએકપણ નહી.
==> $\frac{{dv}}{{dx}} = {v^2} + 1$ ==>$\frac{{dv}}{{{v^2} + 1}} = dx$
On integrating, we get
${\tan ^{ - 1}}v = x + c$ or $v = \tan (x + c)$ ==>$x + y = \tan (x + c)$.
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