MCQ
$\frac{{dy}}{{dx}} = \frac{1}{{x + y + 1}}$ નો ઉકેલ મેળવો.
- ✓$x = c{e^y} - y - 2$
- B$y = x + c{e^y} - 2$
- C$x + c{e^y} - y - 2 = 0$
- Dએકપણ નહી.
==>$\frac{{dx}}{{dy}} - x = y + 1$
It is linear equation, therefore $I.F. $$ = {e^{\int_{}^{} { - 1dy} }} = {e^{ - y}}$
Hence the solution of the equation is
$x.{e^{ - y}} = \int_{}^{} {(y + 1){e^{ - y}}} dy + c$ ==> $x = c{e^y} - y - 2$.
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