The solution of dy dx + y = e^ - x , , ,y(0) = 0, is - Vidyadip

MCQ

The solution of $\frac{{dy}}{{dx}} + y = {e^{ - x}},\,\,y(0) = 0$, is

A
$y = {e^{ - x}}(x - 1)$
B
$y = x{e^x}$
C
$y = x{e^{ - x}} + 1$
✓
$y = x{e^{ - x}}$

✓

Answer

Correct option: D.

$y = x{e^{ - x}}$

d
(d) $\frac{{dy}}{{dx}} + y = {e^{ - x}}$; $I.F.$ $ = {e^{\int {dx} }} = {e^x}$

$y{e^x} = \int {{e^{ - x}}.{e^x}dx + c} $ ==> $y{e^x} = x + c$

Since $y(0) = 0$, $c = 0$

Hence, the required solution is $y{e^x} = x$ ==> $y = x{e^{ - x}}$.

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