Question
Solve the differential equation $\frac{ d y}{ d x}+y= e ^{- x }$
✓
Answer
$\frac{ d y}{ d x}+y= e ^{- x }$
The given equation is of the form
$\frac{ d y}{ d x}+ P y= Q .$
where $P =1$ and $Q = e ^{- x }$
$ \therefore \text { I.F. }= e ^{\iint^{ Pd x}}$
$= e ^{\int d x}$
$= e ^{ X } $
$\therefore$ Solution of the given equation is
$y(\text { I.F. })=\int Q (\text { I.F. }) d x+ c$
$\therefore y \cdot e ^x=\int e ^{-x} \times e ^x d x+ c$
$\therefore y \cdot e ^x=\int e ^{-x+x} d x+ c$
$\therefore y \cdot e ^x=\int e ^0 d x+ c$
$\therefore y \cdot e ^x=\int 1 d x+ c$
$\therefore ye ^{ x }= x + c $
The given equation is of the form
$\frac{ d y}{ d x}+ P y= Q .$
where $P =1$ and $Q = e ^{- x }$
$ \therefore \text { I.F. }= e ^{\iint^{ Pd x}}$
$= e ^{\int d x}$
$= e ^{ X } $
$\therefore$ Solution of the given equation is
$y(\text { I.F. })=\int Q (\text { I.F. }) d x+ c$
$\therefore y \cdot e ^x=\int e ^{-x} \times e ^x d x+ c$
$\therefore y \cdot e ^x=\int e ^{-x+x} d x+ c$
$\therefore y \cdot e ^x=\int e ^0 d x+ c$
$\therefore y \cdot e ^x=\int 1 d x+ c$
$\therefore ye ^{ x }= x + c $
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