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Solve-ydx-xdy-log-xdx-0-

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Question Number 40422 by rahul 19 last updated on 21/Jul/18
Solve: ydx − xdy +log xdx =0
$$\mathrm{Solve}: \\ $$$$\mathrm{yd}{x}\:−\:{xdy}\:+\mathrm{log}\:{xdx}\:=\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 22/Jul/18
⇒y −x(dy/dx) +ln(x)=0 ⇒xy^′ −y −ln(x)=0 ⇒ xy^′ −y =ln(x) he ⇒xy^′ −y =0 ⇒(y^′ /y) =(1/x) ⇒ln∣y∣=ln∣x∣+c ⇒ ⇒y =k∣x∣ case1 x>0 mvc method give y^′ =k^′ x +k e)⇒x(k^′ x+k)−kx =ln(x)⇒ k^′ x^2 =ln(x) ⇒ k^′ =((ln(x))/x^2 ) ⇒ k(x)= ∫_. ^x ((ln(t))/t^2 )dt +c by parts ∫ ((ln(t))/t^2 )dt =−(1/t)ln(t)+∫ (1/t) (dt/t) =−(1/t)ln(t)−(1/t) ⇒k(x)=−((ln(x))/x) −(1/x) +c ⇒ y(x)=x( −((ln(x))/x) −(1/x) +c) =cx−1 −ln(x) case 2 x<0 we follow the same method .
$$\Rightarrow{y}\:\:−{x}\frac{{dy}}{{dx}}\:+{ln}\left({x}\right)=\mathrm{0}\:\Rightarrow{xy}^{'} \:−{y}\:−{ln}\left({x}\right)=\mathrm{0}\:\Rightarrow \\ $$$${xy}^{'} \:−{y}\:\:={ln}\left({x}\right)\:\: \\ $$$${he}\:\Rightarrow{xy}^{'} \:−{y}\:=\mathrm{0}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=\frac{\mathrm{1}}{{x}}\:\Rightarrow{ln}\mid{y}\mid={ln}\mid{x}\mid+{c}\:\Rightarrow \\ $$$$\Rightarrow{y}\:={k}\mid{x}\mid\: \\ $$$${case}\mathrm{1}\:\:{x}>\mathrm{0}\:\:{mvc}\:{method}\:{give}\: \\ $$$${y}^{'} \:={k}^{'} {x}\:+{k} \\ $$$$\left.{e}\right)\Rightarrow{x}\left({k}^{'} {x}+{k}\right)−{kx}\:={ln}\left({x}\right)\Rightarrow \\ $$$${k}^{'} {x}^{\mathrm{2}} \:={ln}\left({x}\right)\:\Rightarrow\:{k}^{'} \:=\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} }\:\Rightarrow\:{k}\left({x}\right)=\:\int_{.} ^{{x}} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt}\:+{c} \\ $$$${by}\:{parts}\:\int\:\:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt}\:=−\frac{\mathrm{1}}{{t}}{ln}\left({t}\right)+\int\:\frac{\mathrm{1}}{{t}}\:\frac{{dt}}{{t}} \\ $$$$=−\frac{\mathrm{1}}{{t}}{ln}\left({t}\right)−\frac{\mathrm{1}}{{t}}\:\Rightarrow{k}\left({x}\right)=−\frac{{ln}\left({x}\right)}{{x}}\:−\frac{\mathrm{1}}{{x}}\:+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)={x}\left(\:\:\:−\frac{{ln}\left({x}\right)}{{x}}\:−\frac{\mathrm{1}}{{x}}\:+{c}\right) \\ $$$$={cx}−\mathrm{1}\:−{ln}\left({x}\right) \\ $$$${case}\:\mathrm{2}\:{x}<\mathrm{0}\:\:{we}\:{follow}\:{the}\:{same}\:{method}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18
((xdy−ydx)/x^2 )=((lnxdx)/x^2 ) d((y/x))=((lnxdx)/x^2 ) ∫d((y/x))=lnx∫x^(−2) dx−∫(1/x).((x^(−1) /(−1)))dx (y/x)=−((lnx)/x)+∫(dx/x^2 )+c (y/x)=−((lnx)/x)−(1/x)+c ((y+1+lnx)/x)=c
$$\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} }=\frac{{lnxdx}}{{x}^{\mathrm{2}} } \\ $$$${d}\left(\frac{{y}}{{x}}\right)=\frac{{lnxdx}}{{x}^{\mathrm{2}} } \\ $$$$\int{d}\left(\frac{{y}}{{x}}\right)={lnx}\int{x}^{−\mathrm{2}} {dx}−\int\frac{\mathrm{1}}{{x}}.\left(\frac{{x}^{−\mathrm{1}} }{−\mathrm{1}}\right){dx} \\ $$$$\frac{{y}}{{x}}=−\frac{{lnx}}{{x}}+\int\frac{{dx}}{{x}^{\mathrm{2}} }+{c} \\ $$$$\frac{{y}}{{x}}=−\frac{{lnx}}{{x}}−\frac{\mathrm{1}}{{x}}+{c} \\ $$$$\frac{{y}+\mathrm{1}+{lnx}}{{x}}={c} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 21/Jul/18
thank you sir ��

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