if dy/dx = 2x+y - 2x/2y, y(0) = 1,


If $\frac{d y}{d x}=\frac{2^{x+y}-2^{x}}{2^{y}}, y(0)=1$, then $y(1)$ is equal to :

  1. $\log _{2}(2+\mathrm{e})$

  2. $\log _{2}(1+\mathrm{e})$

  3. $\log _{2}(2 \mathrm{e})$

  4. $\log _{2}\left(1+\mathrm{e}^{2}\right)$

Correct Option: , 2


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

$2^{y} \frac{d y}{d x}=2^{x}\left(2^{y}-1\right)$

$\int \frac{2^{y}}{2^{y}-1} d y=\int 2^{x} d x$

$\frac{\ln \left(2^{y}-1\right)}{\ln 2}=\frac{2^{x}}{\ln 2}+C$

$\Rightarrow \log _{2}\left(2^{y}-1\right)=2^{x} \log _{2} e+C$

$\because y(0)=1 \Rightarrow 0=\log _{2} e+C$

$\mathrm{C}=-\log _{2} \mathrm{e}$

$\Rightarrow \log _{2}\left(2^{y}-1\right)=\left(2^{x}-1\right) \log _{2} e$

put $x=1, \log _{2}\left(2^{y}-1\right)=\log _{2} e$


$\mathrm{y}=\log _{2}(\mathrm{e}+1)$ Ans.

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