Question

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

• A. ${\text{log}}_2\left(2e\right)$
• B. ${\text{log}}_2(1+e^2)$
• C. ${\text{log}}_2(1+e)$
• D. ${\text{log}}_2(2+e)$

Answer 1

The correct option is C ${\text{log}}_2(1+e)$

$\frac{dy}{dx}=2^x\left(\frac{2^y-1}{2^y}\right)\Rightarrow \frac{2^{y}dy}{2^y-1}=2^{x}dx\Rightarrow \frac{2^y\left(\ln 2\right)dy}{2^y-1}=2^x\left(\ln 2\right)dx$
Integrating on both sides, we get
$\Rightarrow \int \frac{2^y\left(\ln 2\right)dy}{2^y-1}=\int 2^x\left(\ln 2\right)dx\Rightarrow \int \frac{dt}{t}=2^x+c\Rightarrow \ln |t|=2^x+c\Rightarrow \ln |2^y-1|=2^x+c\Rightarrow |2^y-1|=e^{2^x}\cdot e^c\Rightarrow 2^y-1=\pm e^c\cdot e^{2^x}\Rightarrow y={\log }_2(1+k\cdot e^{2^x})$
We know $y\left(0\right)=1$
$1={\log }_2(1+ke)\Rightarrow k=\frac{1}{e}\therefore y\left(1\right)={\log }_2(1+\frac{1}{e}\cdot e^{2^1})=1+{\log }_2(1+e)$