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

$\begin{array}{1 1}(A)\; 2^{-y}-2^{-x}=c \\ (B)\; 2^{-y}+2^{-x}=c \\ (C)\; 2^{y}+2^{x}=c \\(D)\; 2^{-x}-2^{y}=c \end{array}$

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• A relation between the dependent and independent variable which when substituted in the differential equation reduces it to an identify is called a solution.
• Form $\large\frac{dy}{dx}=f(x,y),$ when variable are seperable we express it in the form $f(x)dx=g(y) dy$
• Then $\int g(y)dx=\int f(x) dx+c$
• $\int a^xdx=(a^x/\log\;a)+c$
Given $\large\frac{dy}{dx}$$=2^{y-x} Clearly here the variables can be seperated to the form f(x)dx=g(y)dy (ie) \Large\frac{dy}{dx}=\Large\frac{2^y}{2^x} On seperating the variables we get, \Large\frac{dy}{2^y}=\frac{dx}{2^x} =>2^{-y} dy =2^{-x} dx Integrating on both sides we get \int 2^{-y}dy=\int 2^{-x} dx \int a^xdx=\Large\frac{a^x}{\log\;a}$$+c$
$\Large\frac{2^{-y}}{\log 2}=\frac{2^{-x}}{\log 2}$$+c$
=>$2^{-y}=2^{-x}+c$
or $2^{-y}-2^{-x}=c$