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Question Number 50432 by Abdo msup. last updated on 16/Dec/18

solve x(x^2 −1)y^′ +2y =x^2

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

Commented by Abdo msup. last updated on 16/Dec/18

(he) ⇒x(x^2 −1)y^′ +2y =0 ⇔x(x^2 −1)y^′ =−2y ⇔ (y^′ /y) =((−2)/(x(x^2 −1))) ⇒ln∣y∣=−2 ∫ (dx/(x(x^2 −1))) +c let decompise F(x)=(1/(x(x^2 −1))) ⇒F(x)=(1/(x(x−1)(x+1))) =(a/x) +(b/(x−1)) +(c/(x+1)) a =lim_(x→o) xF(x)=−1 b=lim_(x→1) (x−1)F(x)=(1/2) c=lim_(x→−1) (x+1)F(x)=(1/2) ⇒F(x)=((−1)/x) +(1/(2(x−1))) +(1/(2(x+1))) ⇒ln∣y∣ =−2 ∫(((−1)/x) +(1/(2(x−1))) +(1/(2(x+1))))dx =2ln∣x∣−ln∣x−1∣−ln∣x+1∣ +c ⇒ y (x)=k (x^2 /(∣x^2 −1∣)) let suppose ∣x∣>1 ⇒ y(x) = K (x^2 /(x^2 −1)) mvc method give y^′ = K^′ (x^2 /(x^2 −1)) +K ((2x(x^2 −1)−x^2 (2x))/((x^2 −1)^2 )) =K^′ (x^2 /(x^2 −1)) +K ((−2x)/((x^2 −1)^2 )) and (e) ⇔ x(x^2 −1)(K^′ (x^2 /(x^2 −1)) +K ((−2x)/((x^2 −1)^2 ))) +2K(x^2 /(x^2 −1)) =x^2 ⇒ K^′ x^3 −2x^2 K(1/(x^2 −1)) +2K (x^2 /(x^2 −1)) =x^2 ⇒ K^′ =(1/x) ⇒ K =ln∣x∣ +λ ⇒ y(x) =(x^2 /(x^2 −1))( ln∣x∣ +λ) =((x^2 ln∣x∣)/(x^2 −1)) +((λx^2 )/(x^2 −1)) .

$$\left({he}\right)\:\Rightarrow{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} \:+\mathrm{2}{y}\:=\mathrm{0}\:\Leftrightarrow{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} =−\mathrm{2}{y}\:\Leftrightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=\frac{−\mathrm{2}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow{ln}\mid{y}\mid=−\mathrm{2}\:\int\:\:\:\frac{{dx}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:+{c}\:{let}\:{decompise} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}−\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow{o}} {xF}\left({x}\right)=−\mathrm{1} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({x}\right)=\frac{−\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)} \\ $$$$\Rightarrow{ln}\mid{y}\mid\:=−\mathrm{2}\:\int\left(\frac{−\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\right){dx} \\ $$$$=\mathrm{2}{ln}\mid{x}\mid−{ln}\mid{x}−\mathrm{1}\mid−{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow \\ $$$${y}\:\left({x}\right)={k}\:\frac{{x}^{\mathrm{2}} }{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}\:\:\:{let}\:{suppose}\:\mid{x}\mid>\mathrm{1}\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\:{K}\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:{mvc}\:{method}\:{give} \\ $$$${y}^{'} \:=\:\:{K}^{'} \:\:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:+{K}\:\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)−{x}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={K}^{'} \:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:+{K}\:\frac{−\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:{and}\:\left({e}\right)\:\Leftrightarrow \\ $$$${x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({K}^{'} \:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:+{K}\:\frac{−\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right)\:+\mathrm{2}{K}\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:={x}^{\mathrm{2}} \:\Rightarrow \\ $$$${K}^{'} \:\:{x}^{\mathrm{3}} \:−\mathrm{2}{x}^{\mathrm{2}} {K}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\:+\mathrm{2}{K}\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:={x}^{\mathrm{2}} \:\Rightarrow\:{K}^{'} =\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$$${K}\:={ln}\mid{x}\mid\:+\lambda\:\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\left(\:{ln}\mid{x}\mid\:+\lambda\right)\:=\frac{{x}^{\mathrm{2}} {ln}\mid{x}\mid}{{x}^{\mathrm{2}} \:−\mathrm{1}}\:+\frac{\lambda{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18

(dy/dx)+(2/(x(x^2 −1)))y=(x/(x^2 −1)) I.F e^(∫(2/(x(x^2 −1)))dx) ∫((2x)/(x^2 (x^2 −1)))dx ∫(dt/(t(t−1)))=∫((t−(t−1))/(t(t−1)))dt=∫(dt/(t−1))−∫(dt/t)=ln(((t−1)/t)) I.F e^(ln(((x^2 −1)/x^2 ))) =((x^2 −1)/x^2 ) ((x^2 −1)/x^2 )(dy/dx)+((x^2 −1)/x^2 )×(2/(x(x^2 −1)))y=((x^2 −1)/x^2 )×(x/(x^2 −1)) (1−(1/x^2 ))(dy/dx)+((2/x^3 ))y=(1/x) (d/dx){y×(1−(1/x^2 ))}=(1/x) d{y×(1−(1/x^2 ))}=(dx/x) intregating y×(1−(1/x^2 ))=lnx+c

$$\frac{{dy}}{{dx}}+\frac{\mathrm{2}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{y}=\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${I}.{F}\:\:{e}^{\int\frac{\mathrm{2}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{dx}} \\ $$$$\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}{dx} \\ $$$$\int\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)}=\int\frac{{t}−\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt}=\int\frac{{dt}}{{t}−\mathrm{1}}−\int\frac{{dt}}{{t}}={ln}\left(\frac{{t}−\mathrm{1}}{{t}}\right) \\ $$$${I}.{F}\:\:\:{e}^{{ln}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }\right)} =\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }\frac{{dy}}{{dx}}+\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }×\frac{\mathrm{2}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{y}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }×\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\frac{{dy}}{{dx}}+\left(\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\right){y}=\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{d}}{{dx}}\left\{{y}×\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right\}=\frac{\mathrm{1}}{{x}} \\ $$$${d}\left\{{y}×\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right\}=\frac{{dx}}{{x}} \\ $$$${intregating} \\ $$$${y}×\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)={lnx}+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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