The solution of y' - y = 1, ;y(0) = - 1 is given by y(x) = - Vidyadip

MCQ

The solution of $y' - y = 1,\;y(0) = - 1$ is given by $y(x) = $

A
$ - \exp (x)$
B
$ - \exp ( - x)$
✓
$-1$
D
$\exp (x) - 2$

✓

Answer

Correct option: C.

$-1$

c
(c) $\frac{{dy}}{{dx}} - y = 1$ ==> $\frac{{dy}}{{1 + y}} = dx$

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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