Question Number 82883 by jagoll last updated on 26/Feb/20
$$\left[\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{y}}{\:\sqrt{\mathrm{x}}}\:\right]\:.\frac{\mathrm{dx}}{\mathrm{dy}}\:=\:\mathrm{1}\:,\:\mathrm{x}\:\neq\:\mathrm{0} \\ $$
Answered by john santu last updated on 26/Feb/20
$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}\:−\:\frac{\mathrm{y}}{\:\sqrt{\mathrm{x}}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)\:\mathrm{y}\:=\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}} \\ $$$$\mathrm{IF}\:=\:\mathrm{e}\:^{\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:\mathrm{dx}} \:=\:\mathrm{e}^{\mathrm{2}\sqrt{\mathrm{x}}} \: \\ $$$$\Rightarrow\:\mathrm{y}.\mathrm{e}^{\mathrm{2}\sqrt{\mathrm{x}}} \:\:=\:\int\:\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}\:×\:\mathrm{e}^{\mathrm{2}\sqrt{\mathrm{x}}} \:\mathrm{dx} \\ $$$$\Rightarrow\mathrm{y}.\mathrm{e}^{\mathrm{2}\sqrt{\mathrm{x}}} \:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:\mathrm{dx} \\ $$$$\Rightarrow\:\mathrm{y}.\mathrm{e}^{\mathrm{2}\sqrt{\mathrm{x}}} \:=\:\mathrm{2}\sqrt{\mathrm{x}}\:+\:\mathrm{C} \\ $$$$ \\ $$
Commented by john santu last updated on 26/Feb/20
$$\mathrm{haha}…\mathrm{i}\:\mathrm{will}\:\mathrm{check}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 26/Feb/20
$$\mathrm{soory}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{wrote}\:\mathrm{the}\:\mathrm{problem}\:.\:\mathrm{sir} \\ $$$$\mathrm{john}\:\mathrm{is}\:\mathrm{right}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{w}\:\mathrm{and} \\ $$$$\mathrm{john} \\ $$