If frac{dy}{dx} = frac{2^x y + 2^y cdot 2^x}{ | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{2^x(y + 2^y)}{2^x(1 + 2^y \ln 2)}$.Canceling $2^x$ from the numerator and denominator,we g

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$(1, 2)$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{2^x(y + 2^y)}{2^x(1 + 2^y \ln 2)}$.Canceling $2^x$ from the numerator and denominator,we get $\frac{dy}{dx} = \frac{y + 2^y}{1 + 2^y \ln 2}$.Rearranging the terms to separate variables: $\frac{1 + 2^y \ln 2}{y + 2^y} dy = dx$.Integrating both sides: $\int \frac{1 + 2^y \ln 2}{y + 2^y} dy = \int dx$.Let $u = y + 2^y$,then $du = (1 + 2^y \ln 2) dy$. Thus,$\int \frac{1}{u} du = x + C$.$\ln|y + 2^y| = x + C$.Given $y(0) = 0$,substitute $x = 0$ and $y = 0$: $\ln|0 + 2^0| = 0 + C \Rightarrow \ln(1) = C \Rightarrow C = 0$.So,$x = \ln(y + 2^y)$.For $y = 1$,$x = \ln(1 + 2^1) = \ln(3)$.Since $e \approx 2.718$ and $e^2 \approx 7.389$,and $e Therefore,$x \in (1, 2)$.

If $\frac{dy}{dx} = \frac{2^x y + 2^y \cdot 2^x}{2^x + 2^{x+y} \log_e 2}$ and $y(0) = 0$,then for $y = 1$,the value of $x$ lies in the interval:

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