Question

Find the solution of the differential equation $\left(e^{y-x}\right) d y=\left(e^{x}-e^{y}\right) d x$

• A. $e^{y} e^{x}=e^{2 x}-e^{x^{2}}+c$
• B. $e^{y} e^{x}=e^{x} e^{e^{x}}-e^{e^{x}}+c$
• C. $e^{y} e^{e^{x}}=e^{x} e^{e^{x}}-e^{e^{x}}+c$
• D. $e^{e^{y}} e^{x}=e^{x} e^{e^{x}}-e^{e^{x}}+c$

Answer 1

Answer: (C)

$\left(e^{y-x}\right) d y=\left(e^{x}-e^{y}\right) d x$
$\Rightarrow e^{y} \cdot \frac{d y}{d x}=e^{2 x}-e^{x} \cdot e^{y}$
$\Rightarrow e^{y} \cdot \frac{d y}{d x}+e^{x} \cdot e^{y}=e^{2 x}$
Let
$e^{y}=z \Rightarrow e^{y} \cdot \frac{d y}{d x}=\frac{d z}{d x}$
$\Rightarrow \frac{d z}{d x}+e^{x} \cdot z=e^{2 x}$
IF $=e^{\int e^{z^{x} \cdot d x}}=e^{e^{x}}$
$\therefore $ Solution is
$Z \cdot e^{e^{x}}=\int e^{2 x} \cdot e^{e^{x}} \cdot d x+ c$
Let $e^{x}=u$
$e^{x} \cdot d x=d u$
$\Rightarrow Z \cdot e^{e^{x}}=\int u \cdot e^{u} d u+c=u e^{u}-e^{u}+c$
$\Rightarrow e^{y} \cdot e^{e^{x}}=e^{e^{x}}\left(e^{x}-1\right)+c$