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Question Number 87905 by john santu last updated on 07/Apr/20

$${\mathrm{y}}^{\prime }=2^{\mathrm{y}}$$

Commented by john santu last updated on 07/Apr/20

$$2^{-\mathrm{y}}.{\mathrm{y}}^{\prime }=1$$$${\mathrm{e}}^{-\mathrm{y}\ln 2}.{\mathrm{y}}^{\prime }=1$$$${\left(\frac{{\mathrm{e}}^{-\mathrm{y}\ln 2}}{-\ln 2}\right)}^{,}=1\Rightarrow \frac{{\mathrm{e}}^{-\mathrm{y}\ln 2}}{-\ln 2}=\mathrm{t}+\mathrm{c}$$$${\mathrm{e}}^{-\mathrm{y}\ln 2}=-\mathrm{t}.\ln 2+\mathrm{C}$$$$2^{-\mathrm{y}}=-\mathrm{t}\ln \left(2\right)+\mathrm{C}$$$$\Rightarrow -\mathrm{y}={\log }_2(-\mathrm{t}\ln \left(2\right)+\mathrm{C})$$$$\mathrm{y}=-{\log }_2(\mathrm{C}-\ln \left(2^{\mathrm{t}}\right))$$

Answered by Joel578 last updated on 07/Apr/20

$$\frac{dy}{dx}=2^y\Rightarrow \int 2^{-y}dy=\int dx$$$$\Rightarrow -\frac{2^{-y}}{\ln 2}=x+C_1$$$$\Rightarrow 2^{-y}=(C_2-x)\ln 2,C_2=-C_1$$$$\Rightarrow -y\ln 2=\ln \left[(C_2-x)\ln 2\right]$$$$\Rightarrow y\left(x\right)=-\frac{\ln \left[(C_2-x)\ln 2\right]}{\ln 2}=-{\log }_2\left[(C_2-x)\ln 2\right]$$

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