If [dy dx = 2^xy + 2^y.2^x 2^x + 2^x + y _e2], y(0) = 0, then for y = 1, the value of x li…

Question

If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :

Accepted Answer

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

$$ \Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} } $$

$$ \Rightarrow \ln \left| {y + {2^y}} \right| = x + c$$

x = 0; y = 0 $$\Rightarrow$$ c = 0

$$ \Rightarrow x = \ln \left| {y + {2^y}} \right|$$

$$\Rightarrow$$ at y = 1, x = ln3

$$\because$$ $$3 \in (e,{e^2}) \Rightarrow x \in (1,2)$$

Solutions · 1