"If ${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$, x, y 0, y(1) = 1, then y(2) is equal to :"

Question

Accepted Answer

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${{dy} \over {dx}} + {{{2^{x-y}}({2^y} - 1)} \over {{2^x} - 1}}=0$, x, y > 0, y(1) = 1

${{dy} \over {dx}} = - {{{2^x}({2^y} - 1)} \over {{2^y}({2^x} - 1)}}$

$\int {{{{2^y}} \over {{2^y} - 1}}dy = - \int {{{{2^x}} \over {{2^x} - 1}}dx} }$

$$ = {{{{\log }_e}({2^y} - 1)} \over {{{\log }_e}2}} = - {{{{\log }_e}({2^x} - 1)} \over {{{\log }_e}2}} + {{{{\log }_e}c} \over {{{\log }_e}2}}$$

$= |({2^y} - 1)({2^x} - 1)| = c$

$\because$ $y(1) = 1$

$\therefore$ $c = 1$

$= |({2^y} - 1)({2^x} - 1)| = 1$

For $x = 2$

$|({2^y} - 1)3| = 1$

${2^y} - 1 = {1 \over 3} \Rightarrow 2y = {4 \over 3}$

Taking log to base 2.

$\therefore$ $y = 2 - {\log _2}3$

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If ${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$, x, y > 0, y(1) = 1, then y(2) is equal to :