Question Number 103776 by bemath last updated on 17/Jul/20

y′ − (y/(x^2 −1)) = y^2

Answered by bemath last updated on 17/Jul/20

set v = y^(−1)  ⇒(dv/dx) = −y^(−2)  (dy/dx)  (dy/dx) = −y^2  (dv/dx)  then −y^2  (dv/dx)−(y/(x^2 −1)) = y^2   (dv/dx) + (v/(x^2 −1)) = −1  integrating factor   j(x) = e^(∫ (dx/((x−1)(x+1)))) = e^((1/2)∫(dx/(x−1))−(dx/(x+1)))   j(x)= e^((1/2)ln (((x−1)/(x+1))))  = (√((x−1)/(x+1)))  we get v(x) = ((∫ (√((x−1)/(x+1))).(−1) dx+C)/(√((x−1)/(x+1))))  v(x) =−(√((x+1)/(x−1))) {∫((√(x−1))/(√(x+1))) dx + C}   (1/y) = −(√((x+1)/(x−1))) {∫ ((√(x−1))/(√(x+1))) dx + C }  (1/y) = −(√((x+1)/(x−1))) {(√(x^2 −1))−2tan^(−1) ((√((x+1)/(x−1))))+C}  (1/y)=−x−1+2(√((x+1)/(x−1)))tan^(−1) ((√((x+1)/(x−1))))−C(√((x+1)/(x−1)))

Commented byAr Brandon last updated on 17/Jul/20

What's the secret to knowing the right change of variable to use ? 🙏

Commented bybobhans last updated on 17/Jul/20

(⌢∵⌢) ∫ ((√(x−1))/(√(x+1))) dx = I  I= (√((x−1)(x+1)))− 2tan^(−1) (√((x+1)/(x−1))) + k

Commented bybemath last updated on 17/Jul/20

Bernoulli eq sir

Commented byCoronavirus last updated on 17/Jul/20

it′s a Bernoulli equation

Answered by mathmax by abdo last updated on 17/Jul/20

e⇒y^′ −(1/(x^2 −1))y =y^2  ⇒(y^′ /y^2 )−(1/(x^2 −1))×(1/y) =1  let z =(1/y) ⇒y =(1/z)  ⇒y^′  =−(z^′ /z^2 )  e⇒−(z^′ /z^2 )×z^2 −(1/(x^2 −1))×z =1 ⇒−z^′ −(1/(x^2 −1))z =1 ⇒z^′ +(z/(x^2 −1)) =−1  e^,   he→z^′  =−(z/(x^2 −1)) ⇒(z^′ /z) =−(1/(x^2 −1)) ⇒ln∣z∣ =−∫ (dx/(x^2 −1)) =−(1/2)∫((1/(x−1))−(1/(x+1)))dx  =−(1/2)ln∣((x−1)/(x+1))∣ +c  =(1/2)ln∣((x+1)/(x−1))∣ +c ⇒z =k (√(∣((x+1)/(x−1))∣))  solution  on w ={x/((x+1)/(x−1))>0} mvc method →z^′  =k^′ (√((x+1)/(x−1)))  +k ×(((((x+1)/(x−1)))^′ )/(2(√((x+1)/(x−1))))) =k^′ (√((x+1)/(x−1)))+(k/2)×((−2)/((x−1)^2 (√((x+1)/(x−1)))))  e^′  ⇒ k^′ (√((x+1)/(x−1)))−(k/((x−1)^2 (√((x+1)/(x−1))))) +(1/(x^2 −1))×k(√(((x+1)/(x−1)) ))=−1 ⇒  k^′  =−(√((x−1)/(x+1))) ⇒k =−∫ (√((x−1)/(x+1)))dx  changement (√((x−1)/(x+1)))=t give  ((x−1)/(x+1))=t^2  ⇒x−1 =t^2 x+t^2  ⇒(1−t^2 )x =t^2  +1 ⇒x =((t^2  +1)/(1−t^2 ))  dx =((2t(1−t^2 )−(t^2  +1)(−2t))/((1−t^2 )^2 ))dt =((2t−2t^3 +2t^3  +2t)/((t^2 −1)^2 ))dt =((4tdt)/((t^2 −1)^2 )) ⇒  ∫ (√((x−1)/(x+1)))dx =4 ∫ ((tdt)/((t^2 −1)^2 )) =4 ∫  ((tdt)/((t−1)^2 (t+1)^2 ))  =4 ∫  ((tdt)/((((t−1)/(t+1)))^2 (t+1)^4 ))  we do the changement ((t−1)/(t+1))=z ⇒t−1 =zt+z ⇒  (1−z)t=1+z ⇒t =((1+z)/(1−z)) ⇒(dt/dz) =((1−z+(1+z))/((1−z)^2 )) =(2/((1−z)^2 ))  t+1 =((1+z)/(1−z)) +1 =((1+z+1−z)/(1−z)) =(2/(1−z)) ⇒I= ∫ ((1+z)/(1−z))×(2/((1−z)^2 z^2 ((2/(1−z)))^4 ))dz  =(1/8)∫    (((1+z)(1−z)^4 )/((1−z)^3 ))dz =(1/8)∫ (1−z^2 )dz  =(1/8)(z−(z^3 /3)) +c =(1/8)(((t−1)/(t+1)))−(1/(24))(((t−1)/(t+1)))^3  +c  with t =(√((x−1)/(x+1)))  y(x)=(1/(z(x)))