The solution of frac{dy}{dx} = 2^{y - x} is | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = 2^{y - x}$.Using the property of exponents,we can write this as $\frac{dy}{dx} = \frac{2^y}{2^x}$

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$\frac{1}{2^x} - \frac{1}{2^y} = c$

Vedclass · Answer

(C) Given the differential equation $\frac{dy}{dx} = 2^{y - x}$.Using the property of exponents,we can write this as $\frac{dy}{dx} = \frac{2^y}{2^x}$.Separating the variables,we get $\frac{dy}{2^y} = \frac{dx}{2^x}$,which is $2^{-y} dy = 2^{-x} dx$.Integrating both sides,we have $\int 2^{-y} dy = \int 2^{-x} dx$.Using the formula $\int a^u du = \frac{a^u}{\ln a}$,we get $\frac{2^{-y}}{-\ln 2} = \frac{2^{-x}}{-\ln 2} + C_1$.Multiplying by $-\ln 2$,we get $2^{-y} = 2^{-x} - C_1 \ln 2$.Rearranging the terms,$2^{-x} - 2^{-y} = C_1 \ln 2$.Let $C = C_1 \ln 2$,then $\frac{1}{2^x} - \frac{1}{2^y} = c$.

The solution of $\frac{dy}{dx} = 2^{y - x}$ is

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