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

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Aicharya · Accepted Answer

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

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

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