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

Given that $\frac{{dy}}{{dx}} = {2^{y - x}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{{2^y}}}{{{2^x}}}$

$\Rightarrow$ $\frac{{dy}}{{{2^y}}} = \frac{{dx}}{{{2^x}}}$

On integrationg both sides,

we get

$\int {{2^{ - y}}} dy = \int {{2^{ - x}}} dx$

$\Rightarrow$ $\frac{{ - {2^{ - y}}}}{{\log 2}} = \frac{{ - {2^{ - x}}}}{{\log 2}} + C$

$\Rightarrow$ $- {2^{ - y}} + {2^{ - x}} = + C\log 2$

$\Rightarrow$ ${2^{ - x}} - {2^{ - y}} = - C\log 2$

$\Rightarrow$ ${2^{ - x}} - {2^{ - y}} = K$

[where, $K = + C\log 2$]