Question
Find the solution of $\frac{\text{dy}}{\text{dx}}=2^\text{y-x}.$

Answer

Given that, $\frac{\text{dy}}{\text{dx}}=2^\text{y-x}$
$\Big[\because\text{a}^\text{m-n}=\frac{\text{a}^\text{m}}{\text{a}^\text{n}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2^\text{y}}{2^\text{x}}$
$\Rightarrow\frac{\text{dy}}{2^\text{y}}=\frac{\text{dx}}{2^\text{x}}$
On integrating both sides, we get
$\int2^\text{-y}\text{dy}=\int2^\text{x}\text{dx}$
$\Rightarrow\frac{-2^\text{-y}}{\log2}=\frac{-2^\text{-x}}{\log2}+\text{C}$
$\Rightarrow-2^\text{-y}+2^\text{-x}=+\text{C}\log2$
$\Rightarrow-2^\text{-x}+2^\text{-x}=+\text{C}\log2$
$\Rightarrow2^\text{-x}-2^\text{-y}=-\text{C}\log2$
$\Rightarrow2^\text{-x}-2^\text{-y}=\text{K}$ $[\text{where}, \text{K} = +\text{C}\log2]$

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