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e-2-x-x-y-x-dx-dy-1-x-0-

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

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