MCQ
$y' - y = 1,\;y(0) = - 1$ નો ઉકેલ $y(x) = $
- A$ - \exp (x)$
- B$ - \exp ( - x)$
- ✓$-1$
- D$\exp (x) - 2$
Integrating both sides $\log (1 + y) = x + c$ ==> $1 + y = {e^x}.{e^c}$
$\because x=0,\,\,y=-1$.Then, $1 - 1 = e.{e^c}$ ==>${e^c} = 0$
Therefore solution $1 + y = {e^x} \times 0 \Rightarrow y(x) = - 1$.
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