If dydx+2^x-y(2^y-1)2^x-1=0, x,y0, y(1)=1, then y(2) equals:

Question

If $\dfrac{dy}{dx}+\dfrac{2^{x-y}(2^y-1)}{2^{x}-1}=0$, $x,y>0$, $y(1)=1$, then $y(2)$ equals:

Accepted Answer

Separating variables: $\displaystyle\int\frac{2^y}{2^y-1}\,dy = -\int\frac{2^x}{2^x-1}\,dx$, giving
$$\log_2(2^y-1)+\log_2(2^x-1) = \log_2 c \implies (2^y-1)(2^x-1)=c.$$

Using $y(1)=1$: $c=1$. At $x=2$: $(2^y-1)(3)=1 \Rightarrow 2^y=\dfrac43 \Rightarrow y=2-\log_23$.

Solutions · 1