If dydx=2^x+y-2x^22^x2^y, y(0)=1, then y(1) is:

Question

If $\dfrac{dy}{dx}=\dfrac{2^{x+y}-2x^2\cdot2^x}{2^y}$, $y(0)=1$, then $y(1)$ is:

Accepted Answer

$$\int \frac{2^y}{2^y-1}\,dy = \int 2^x\,dx.$$

$$\frac{\ln(2^y-1)}{\ln2} = \frac{2^x}{\ln2}+C.$$

Using $y(0)=1$: $0 = \dfrac{1}{\ln2}+C \implies$ at $y(1)$: $\ln2(2^y-1)=2^1-1=1 \implies 2^y-1=e \implies y=\log_2(1+e)$.

Solutions · 1