Question Number 108406 by pticantor last updated on 16/Aug/20
$$ \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{differential}} \\ $$$$\boldsymbol{{equation}}\: \\ $$$$\boldsymbol{{y}}^{'} −\mathrm{2}{e}^{{x}} \boldsymbol{{y}}=\mathrm{2}{e}^{{x}} \sqrt{\boldsymbol{{y}}} \\ $$
Answered by Dwaipayan Shikari last updated on 16/Aug/20
$$\frac{{dy}}{{dx}}=\mathrm{2}{e}^{{x}} \left({y}+\sqrt{{y}}\right) \\ $$$$\int\frac{{dy}}{{y}+\sqrt{{y}}}=\int\mathrm{2}{e}^{{x}} {dx} \\ $$$$\int\:\frac{{dy}}{\:\sqrt{{y}}\left(\mathrm{1}+\sqrt{{y}}\right)}{dy}=\mathrm{2}{e}^{{x}} +{C}\:\:\:\:\:\:\:\:\: \\ $$$$\int\frac{\mathrm{2}{t}}{{t}\left(\mathrm{1}+{t}\right)}{dt}=\mathrm{2}{e}^{{x}} +{C}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}={t}^{\mathrm{2}} ,\:\:\mathrm{1}=\mathrm{2}{t}\frac{{dt}}{{dy}} \\ $$$$\mathrm{2}{log}\left(\mathrm{1}+{t}\right)=\mathrm{2}{e}^{{x}} +{C} \\ $$$${log}\left(\mathrm{1}+{t}\right)={e}^{{x}} +\frac{{C}}{\mathrm{2}} \\ $$$${log}\left(\mathrm{1}+\sqrt{{y}}\right)={e}^{{x}} +\frac{{C}}{\mathrm{2}} \\ $$