The particular solution of d y d x =x e^ y-x , when x = y =0 is..… - Vidyadip

MCQ

The particular solution of $\frac{d y}{d x}=x e^{y-x}$, when $x = y =0$ is......

✓
$e^{x-y}=x+1$
B
$e^{x+y}=x+1$
C
$e^x+e^y=x+1$
D
$e^{y-x}=x-1$

✓

Answer

Correct option: A.

$e^{x-y}=x+1$

$e^{x-y}=x+1$
Hint :
$\frac{d y}{d x}=x e^{y-x}=\frac{x e^y}{e^x}$
$\therefore \int e^{-y} d y=\int x e^{-x} d x$
$\therefore e^{-y}=x \cdot \frac{e^{-x}}{-1}-\int 1 \cdot \frac{e^{-x}}{-1} d x+c$
$\therefore-e^{-y}=-x e^{-x}+\frac{e^{-x}}{-1}+c$
$\therefore e^{-y}=\frac{x}{e^x}+\frac{1}{e^x}-c $
$ \therefore e^{x-y}=x+1-c e^x$
When $x=y=0$, we get
$1=1-c \quad \therefore c=0$
$\therefore$ particular solution is
$\left.e^{x-y}=x+1\right]$

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