Question
Find the solution of $\frac{d y}{d x}=2^{y-x}$
✓
Answer
Given that,$\frac{d y}{d x}=2^{y-x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{2^y}{2^x}$ $\left[\because a^{m-n}=\frac{a^m}{a^n}\right]$
$\Rightarrow \quad \frac{d y}{2^y}=\frac{d y}{2^x}$
On integrating both sides,we get
$\int 2^{-y} d y=\int 2^{-x} d x$
$\Rightarrow \quad \frac{-2^{-y}}{\log 2}=\frac{-2^{-x}}{\log 2}+C$
$\begin{array}{l}\Rightarrow-2^{-y}+2^{-x}=+C \log 2 \\ \Rightarrow \quad 2^{-x}-2^{-y}=-C \log 2\end{array}$
$\Rightarrow \quad 2^{-x}-2^{-y}=k \quad[$ where, $k=+C \log 2]$
$\Rightarrow \quad \frac{d y}{d x}=\frac{2^y}{2^x}$ $\left[\because a^{m-n}=\frac{a^m}{a^n}\right]$
$\Rightarrow \quad \frac{d y}{2^y}=\frac{d y}{2^x}$
On integrating both sides,we get
$\int 2^{-y} d y=\int 2^{-x} d x$
$\Rightarrow \quad \frac{-2^{-y}}{\log 2}=\frac{-2^{-x}}{\log 2}+C$
$\begin{array}{l}\Rightarrow-2^{-y}+2^{-x}=+C \log 2 \\ \Rightarrow \quad 2^{-x}-2^{-y}=-C \log 2\end{array}$
$\Rightarrow \quad 2^{-x}-2^{-y}=k \quad[$ where, $k=+C \log 2]$
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