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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
The solution of $\frac{{dy}}{{dx}} = {2^{y - x}}$ is
  • A
    ${2^x} + {2^y} = c$
  • B
    ${2^x} - {2^y} = c$
  • ✓
    $\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = c$
  • D
    $x + y = c$
✓

Answer

Correct option: C.
$\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = c$
c
(c) Given $\frac{{dy}}{{dx}} = {2^{y - x}}$$ = \frac{{{2^y}}}{{{2^x}}}$ or $\frac{{dy}}{{{2^y}}} = \frac{{dx}}{{{2^x}}}$

Integrating both sides, $\int {\frac{{dy}}{{{2^y}}} = \int {\frac{{dx}}{{{2^x}}}} } $
$ - {2^{ - y}}\log 2 = - {2^{ - x}}\log 2 + {c_1}$

$\frac{{\log 2}}{{{2^x}}} - \frac{{\log 2}}{{{2^y}}} = {c_1}$; $\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = \frac{{{c_1}}}{{\log 2}} = c$.

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