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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
Solution of the equation $({e^x} + 1)ydy = (y + 1){e^x}dx$ is
  • A
    $c(y + 1)({e^x} + 1) + {e^y} = 0$
  • B
    $c(y + 1)({e^x} - 1) + {e^y} = 0$
  • C
    $c(y + 1)({e^x} - 1) - {e^y} = 0$
  • ✓
    $c(y + 1)({e^x} + 1) = {e^y}$
✓

Answer

Correct option: D.
$c(y + 1)({e^x} + 1) = {e^y}$
d
(d) $({e^x} + 1)ydy = (y + 1){e^x}dx$

==> $\left( {\frac{y}{{y + 1}}} \right)dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)\,dx$ ==> $\left[ {1 - \frac{1}{{y + 1}}} \right]dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)\,dx$

==> $\int_{}^{} {\left\{ {1 - \frac{1}{{y + 1}}} \right\}} dy = \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}} dx$

==> $y = \log (y + 1) + \log ({e^x} + 1) + \log c$ or ${e^y} = c(y + 1)({e^x} + 1)$.

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