Question
The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is
✓
Answer
We have, $ \frac{d y}{d x}=e^{x+y}$
$\Rightarrow \frac{d y}{d x}=e^{x} \times e^{y}$
separating variables
$\Rightarrow \mathrm{e}^{-\mathrm{y}} \mathrm{dy}=\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
Integrating both sides
$\Rightarrow \int \mathrm{e}^{-\mathrm{y}} \mathrm{d} \mathrm{y}=\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
$\Rightarrow-e^{-y}=e^{x}+c$
$\Rightarrow e^{x}+e^{-y}=-c$
Or,
$e^{x}+e^{-y}=c$ (c is a constant)
$\Rightarrow \frac{d y}{d x}=e^{x} \times e^{y}$
separating variables
$\Rightarrow \mathrm{e}^{-\mathrm{y}} \mathrm{dy}=\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
Integrating both sides
$\Rightarrow \int \mathrm{e}^{-\mathrm{y}} \mathrm{d} \mathrm{y}=\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
$\Rightarrow-e^{-y}=e^{x}+c$
$\Rightarrow e^{x}+e^{-y}=-c$
Or,
$e^{x}+e^{-y}=c$ (c is a constant)
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