Question

If $\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y},y\left(0\right)=1,$ then which of the following is/are true

• A. $y\left(1\right)={\log }_2(1+e)$
• B. $y\left(1\right)={\log }_2(1+e^2)$
• C. $y\left(2\right)={\log }_2(1+e^3)$
• D. $y\left(2\right)={\log }_2(1+e^4)$

Answer 1

Answer: (A), (C)

$y\left(2\right)={\log }_2(1+e^3)$

Given : $\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y},y\left(0\right)=1,$
$\Rightarrow \frac{dy}{dx}=2^x\left(\frac{2^y-1}{2^y}\right)$
$\Rightarrow \left(\frac{2^y}{2^y-1}\right)dy=2^{x}dx$
Integrating both sides, we get
$\int \left(\frac{2^y}{2^y-1}\right)dy=\int 2^{x}dx$
Let $2^y-1=t\Rightarrow 2^y\ln 2dy=dt$
$\Rightarrow \int \frac{dt}{t}\left(\frac{1}{\ln 2}\right)=\int 2^{x}dx$
$\Rightarrow \frac{\ln t}{\ln 2}=\frac{2^x}{\ln 2}+C$
$\therefore \ln (2^y-1)=2^x+c$
When $x=0,y=1\Rightarrow c=-1$
$\Rightarrow 2^y=1+\left(e^{2^x-1}\right)$
When $x=1,2^y=1+e$
$\therefore y\left(1\right)={\log }_2(1+e)$
When $x=2,2^y=1+e^3$
$\therefore y\left(2\right)={\log }_2(1+e^3)$