Question

Solution of the differential eqaution $(1+\ln 2-\frac{dy}{dx})2^x=\frac{dy}{dx}-1$ if the curve passes through $(0,\ln 2)$ is

• A. $y=\ln (1+2^x)-x$
• B. $y=\ln (1+2^x)+x$
• C. $y=\ln (2^x-1)+x$
• D. $y=\ln (2^x-1)-x$

Answer 1

The correct option is B $y=\ln (1+2^x)+x$

Differential equation
$(1+\ln 2-\frac{dy}{dx})2^x=\frac{dy}{dx}-1(1+\ln 2)2^x-2^x\frac{dy}{dx}=\frac{dy}{dx}-12^x\frac{dy}{dx}+\frac{dy}{dx}=(1+\ln 2)2^x+1(2^x+1)\frac{dy}{dx}=2^x\ln 2+1+2^x\frac{dy}{dx}=\frac{2^x\ln 2}{2^x+1}+1dy=\frac{2^x\ln 2}{2^x+1}dx+1dx$
Integrating both the sides
$\int 1dy=\int \frac{2^x\ln 2}{2^x+1}dx+\int 1dx$

Let $2^x+1=y\Rightarrow dy=2^x\ln 2$
$\therefore y=\int \frac{1}{t}dt+xy=x+\ln t+cy=x+\ln (1+2^x)+c$
Now, the curve passes through $(0,\ln 2)$
putting the values in the equation of the curve
$\ln 2=0+\ln (1+1)+c\Rightarrow c=0\therefore y=x+\ln (1+2^x)$