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Question Number 57415 by Abdo msup. last updated on 03/Apr/19
solve (x−1)y^′ +(1+(√x))y =x e^(−2x)
solve(x−1)y′+(1+x)y=xe−2x
Commented by maxmathsup by imad last updated on 12/Apr/19
due to (√x)we must have x≥0 (ed) ⇔((√x)−1)((√x)+1)y^′ +((√( x))+1)y =xe^(−2x) ⇒ ((√x)−1)y^′ +y =(x/( (√x)+1)) e^(−2x) (e) (he) ⇒((√x)−1)y^′ +y =0 ⇒((√x)−1)y^′ =−y ⇒(y^′ /y) =−(1/( (√x)−1)) ⇒ ln∣y∣ =−∫ (dx/( (√x)−1)) =_((√x)=t) −∫ ((2tdt)/(t−1)) =−2 ∫ ((t−1+1)/(t−1)) dt =−2t −2ln∣t−1∣ +c ⇒ y(x) =(K/((t−1)^2 )) e^(−2t) =(K/(((√x)−1)^2 )) e^(−2(√x)) let use mvc method we have y^′ (x) =K^′ ((√x)−1)^(−2) e^(−2(√x)) +K { −2 (1/(2(√x)))((√x)−1)^(−3) e^(−2(√x)) −2 (1/(2(√x))) ((√x)−1)^(−2) e^(−2(√x)) } =K^′ ((√x)−1)^(−2) e^(−2(√x)) −(K/( (√x))){ (1/(((√x)−1)^3 )) +(1/(((√x)−1)^2 ))}e^(−2(√x)) and (e) ⇒ ((√x)−1)(K^′ /(((√x)−1)^2 ))−(K/( (√x))){ (1/(((√x)−1)^2 )) +(1/( (√x)−1))} +(K/(((√x)−1)^2 )) =(x/( (√x) +1)) ⇒ (K^′ /( (√x)−1)) −(K/( (√x))){((√x)/(((√x)−1)^2 ))} +(K/(((√x)−1)^2 )) =(x/( (√x) +1)) ⇒K^′ =((x((√x)−1))/( (√x)+1)) ⇒ K(x) =∫ ((x((√x)−1))/( (√x)+1)) dx +c but ∫ ((x((√x)−1))/( (√x)+1)) dx =_((√x)=t) ∫ ((t^2 (t−1))/(t+1))(2t)dt =2 ∫ ((t^3 −t^2 )/(t+1)) dt =2 ∫ ((t^3 +1 −t^2 −1)/(t+1)) dt =2 ∫ (t^2 −t+1) −2 ∫ ((t^2 +1)/(t+1)) dt =(2/3)t^3 −t^2 +2t −2 ∫ ((t^2 −1 +2)/(t+1)) dt =(2/3)t^3 −t^2 +2t −2 ∫(t−1)dt −4ln∣t+1∣ =(2/3)t^3 −2t^2 +4t −4ln∣t+1∣ =(2/3)((√x))^3 −2x +4(√x) −4ln∣1+(√x)∣ ⇒ K(x) =((2x)/3)(√x)−2x+4(√x) −4ln(1+(√x))+c ⇒ y(x) =(e^(−2(√x)) /(((√x)−1)^2 )){ (2/3)x(√x)−2x +4(√x)−4ln(1+(√x)) +C} .
duetoxwemusthavex⩾0(ed)⇔(x−1)(x+1)y′+(x+1)y=xe−2x⇒(x−1)y′+y=xx+1e−2x(e)(he)⇒(x−1)y′+y=0⇒(x−1)y′=−y⇒y′y=−1x−1⇒ln∣y∣=−∫dxx−1=x=t−∫2tdtt−1=−2∫t−1+1t−1dt=−2t−2ln∣t−1∣+c⇒y(x)=K(t−1)2e−2t=K(x−1)2e−2xletusemvcmethodwehavey′(x)=K′(x−1)−2e−2x+K{−212x(x−1)−3e−2x−212x(x−1)−2e−2x}=K′(x−1)−2e−2x−Kx{1(x−1)3+1(x−1)2}e−2xand(e)⇒(x−1)K′(x−1)2−Kx{1(x−1)2+1x−1}+K(x−1)2=xx+1⇒K′x−1−Kx{x(x−1)2}+K(x−1)2=xx+1⇒K′=x(x−1)x+1⇒K(x)=∫x(x−1)x+1dx+cbut∫x(x−1)x+1dx=x=t∫t2(t−1)t+1(2t)dt=2∫t3−t2t+1dt=2∫t3+1−t2−1t+1dt=2∫(t2−t+1)−2∫t2+1t+1dt=23t3−t2+2t−2∫t2−1+2t+1dt=23t3−t2+2t−2∫(t−1)dt−4ln∣t+1∣=23t3−2t2+4t−4ln∣t+1∣=23(x)3−2x+4x−4ln∣1+x∣⇒K(x)=2x3x−2x+4x−4ln(1+x)+c⇒y(x)=e−2x(x−1)2{23xx−2x+4x−4ln(1+x)+C}.

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