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9. differential equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths9. differential equations1 Mark
MCQ
The solution of the equation $\frac{{dy}}{{dx}} = \frac{1}{{x + y + 1}}$ is
  • ✓
    $x = c{e^y} - y - 2$
  • B
    $y = x + c{e^y} - 2$
  • C
    $x + c{e^y} - y - 2 = 0$
  • D
    None of these
✓

Answer

Correct option: A.
$x = c{e^y} - y - 2$
a
(a) $\frac{{dy}}{{dx}} = \frac{1}{{x + y + 1}}$==> $\frac{{dx}}{{dy}} = x + y + 1$

==>$\frac{{dx}}{{dy}} - x = y + 1$

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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