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Question Number 90968 by abdomathmax last updated on 27/Apr/20

solve (1+e^x )y^′ −y =(e^x /(1+x^2 ))

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

Commented by niroj last updated on 27/Apr/20

(1+e^x )y^′ −y= (e^x /(1+x^2 )) (dy/dx)−(1/(1+e^x ))y = (e^x /((1+e^x )(1+x^2 ))) P= −(1/(1+e^x )) , Q= (e^x /((1+e^x )(1+x^2 ))) IF= e^(∫Pdx) = e^(−∫(1/(1+e^x ))dx) −∫ (1/(e^x (e^(−x) +1)))dx −∫(( e^(−x) dx)/(e^(−x) +1)) put e^(−x) +1=t −e^(−x) dx=dt ∫ (1/t)dt=logt =log (e^(−x) +1) IF= e^(log(e^(−x) +1)) = e^(−x) +1 we know, y.IF=∫IF.Qdx+C y.(e^(−x) +1)= ∫ (e^(−x) +1).(e^x /((1+e^x )(1+x^2 )))dx+C y ((1/e^x )+1)=∫ (((1+e^x ))/e^x ).(e^x /((1+e^x )(1+x^2 )))dx+C ((y(1+e^x ))/e^x )=∫(( 1)/(1+x^2 ))dx +C y= ((e^x (tan^(−1) x+C))/((1+e^x )))//.

$$\:\:\:\:\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\mathrm{y}^{'} −\mathrm{y}=\:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\mathrm{y}\:=\:\frac{\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$\:\:\mathrm{P}=\:−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\:,\:\mathrm{Q}=\:\frac{\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$\:\:\:\mathrm{IF}=\:\mathrm{e}^{\int\mathrm{Pdx}} =\:\mathrm{e}^{−\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\mathrm{dx}} \\ $$$$\:\:\:−\int\:\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{x}} \left(\mathrm{e}^{−\mathrm{x}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$\:\:\:\:−\int\frac{\:\:\mathrm{e}^{−\mathrm{x}} \mathrm{dx}}{\mathrm{e}^{−\mathrm{x}} +\mathrm{1}} \\ $$$$\:\:\mathrm{put}\:\mathrm{e}^{−\mathrm{x}} +\mathrm{1}=\mathrm{t} \\ $$$$\:\:\:\:−\mathrm{e}^{−\mathrm{x}} \mathrm{dx}=\mathrm{dt} \\ $$$$\:\:\int\:\frac{\mathrm{1}}{\mathrm{t}}\mathrm{dt}=\mathrm{logt}\:=\mathrm{log}\:\left(\mathrm{e}^{−\mathrm{x}} +\mathrm{1}\right) \\ $$$$\:\:\mathrm{IF}=\:\mathrm{e}^{\mathrm{log}\left(\mathrm{e}^{−\mathrm{x}} +\mathrm{1}\right)} =\:\mathrm{e}^{−\mathrm{x}} +\mathrm{1} \\ $$$$\:\:\mathrm{we}\:\mathrm{know}, \\ $$$$\:\:\mathrm{y}.\mathrm{IF}=\int\mathrm{IF}.\mathrm{Qdx}+\mathrm{C} \\ $$$$\:\:\:\mathrm{y}.\left(\mathrm{e}^{−\mathrm{x}} +\mathrm{1}\right)=\:\int\:\left(\mathrm{e}^{−\mathrm{x}} +\mathrm{1}\right).\frac{\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}+\mathrm{C} \\ $$$$\:\mathrm{y}\:\left(\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{x}} }+\mathrm{1}\right)=\int\:\frac{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)}{\mathrm{e}^{\mathrm{x}} }.\frac{\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}+\mathrm{C} \\ $$$$\:\frac{\mathrm{y}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)}{\mathrm{e}^{\mathrm{x}} }=\int\frac{\:\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:+\mathrm{C} \\ $$$$\:\mathrm{y}=\:\frac{\mathrm{e}^{\mathrm{x}} \left(\mathrm{tan}^{−\mathrm{1}} \mathrm{x}+\mathrm{C}\right)}{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)}//. \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 28/Apr/20

(he)→(1+e^x )y^′ −y =0 ⇒(y^′ /y) =(1/(1+e^x )) ⇒ln∣y∣=∫ (dx/(1+e^x )) +c =_(e^x =t) ∫ (dt/(t(1+t))) =∫((1/t)−(1/(1+t)))dt +c =ln∣(t/(1+t))∣ +c =ln∣(e^x /(1+e^x ))∣ +c y(x)=k×(e^x /(1+e^x )) let use mvc method →y^′ =k^′ (e^x /(1+e^x )) +k×(e^x /((1+e^x )^2 )) (e)⇒k^′ e^x +((ke^x )/(1+e^x ))−((ke^x )/(1+e^x )) =(e^x /(1+x^2 )) ⇒k^′ =(1/(1+x^2 )) ⇒k(x) =arctanx +c the general solution is y(x)=(e^x /(1+e^x ))k(x) =(e^x /(1+e^x ))(arctanx +c)

$$\left({he}\right)\rightarrow\left(\mathrm{1}+{e}^{{x}} \right){y}^{'} −{y}\:=\mathrm{0}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=\frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\:\Rightarrow{ln}\mid{y}\mid=\int\:\:\frac{{dx}}{\mathrm{1}+{e}^{{x}} }\:+{c} \\ $$$$=_{{e}^{{x}} ={t}} \:\:\int\:\:\frac{{dt}}{{t}\left(\mathrm{1}+{t}\right)}\:=\int\left(\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt}\:+{c}\:={ln}\mid\frac{{t}}{\mathrm{1}+{t}}\mid\:+{c}\:={ln}\mid\frac{{e}^{{x}} }{\mathrm{1}+{e}^{{x}} }\mid\:+{c} \\ $$$${y}\left({x}\right)={k}×\frac{{e}^{{x}} }{\mathrm{1}+{e}^{{x}} }\:{let}\:{use}\:{mvc}\:{method}\:\rightarrow{y}^{'} ={k}^{'} \:\frac{{e}^{{x}} }{\mathrm{1}+{e}^{{x}} } \\ $$$$+{k}×\frac{{e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} } \\ $$$$\left({e}\right)\Rightarrow{k}^{'} \:{e}^{{x}} \:\:+\frac{{ke}^{{x}} }{\mathrm{1}+{e}^{{x}} }−\frac{{ke}^{{x}} }{\mathrm{1}+{e}^{{x}} }\:=\frac{{e}^{{x}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow{k}^{'} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow{k}\left({x}\right)\:={arctanx}\:+{c} \\ $$$${the}\:{general}\:{solution}\:{is} \\ $$$${y}\left({x}\right)=\frac{{e}^{{x}} }{\mathrm{1}+{e}^{{x}} }{k}\left({x}\right)\:=\frac{{e}^{{x}} }{\mathrm{1}+{e}^{{x}} }\left({arctanx}\:+{c}\right) \\ $$

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