If [dy dx + 2^x - y(2^y - 1) 2^x - 1 = 0], x, y 0, y(1) = 1, then y(2) is equal to :

Question

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 :

Accepted Answer

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

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

Solutions · 1