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Question Number 115539 by bemath last updated on 26/Sep/20

(dy/dx) = ((e^(tan^(−1) (x)) −y)/(1+x^2 ))

$$\frac{{dy}}{{dx}}\:=\:\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:−{y}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Answered by bobhans last updated on 26/Sep/20

(dy/dx) + (y/(1+x^2 )) = (e^(tan^(−1) (x)) /(1+x^2 )) Integrating factor IF = e^(∫ (dx/(1+x^2 ))) = e^(tan^(−1) (x)) solution : y.e^(tan^(−1) (x)) = ∫ e^(tan^(−1) (x)) .(e^(tan^(−1) (x)) /(1+x^2 )) dx y.e^(tan^(−1) (x)) = ∫ (e^(2tan^(−1) (x)) /(1+x^2 )) dx set tan^(−1) (x)= u ⇒x = tan u ∫ (e^(2u) /(1+tan^2 u)) sec^2 u du = (1/2)e^(2u) +c ∴ y.e^(tan^(−1) (x)) = (1/2)e^(2tan^(−1) (x)) + c y = (1/2)e^(tan^(−1) (x)) + C.e^(−tan^(−1) (x)) .

$$\frac{{dy}}{{dx}}\:+\:\frac{{y}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${Integrating}\:{factor}\: \\ $$$${IF}\:=\:{e}^{\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }} \:=\:{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \\ $$$${solution}\::\:{y}.{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:=\:\int\:{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:.\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$${y}.{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:=\:\int\:\frac{{e}^{\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\: \\ $$$${set}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\:{u}\:\Rightarrow{x}\:=\:\mathrm{tan}\:{u} \\ $$$$\int\:\frac{{e}^{\mathrm{2}{u}} }{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {u}}\:\mathrm{sec}\:^{\mathrm{2}} {u}\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{u}} +{c} \\ $$$$\therefore\:{y}.{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)} \:+\:{c}\: \\ $$$${y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:+\:{C}.{e}^{−\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:. \\ $$

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