Question Number 104857 by ~blr237~ last updated on 24/Jul/20

  y′=((y−x)/(y+x))       y(0)=0

Answered by bemath last updated on 24/Jul/20

y=px ⇒ (dy/dx) = p+x(dp/dx)  ⇔p+x(dp/dx) = ((x(p−1))/(x(p+1)))  x(dp/dx) = ((p−1)/(p+1))−((p(p+1))/(p+1))  x(dp/dx) = ((−p^2 −1)/(p+1)) ⇒((p+1)/(p^2 +1))dp=−(dx/x)  (1/2)∫((d(p^2 +1))/(p^2 +1))+∫(dp/(p^2 +1))= −lnx +C  (1/2)ln (p^2 +1)+tan^(−1) (p)=−lnx+C  ln (x(√(p^2 +1)))+tan^(−1) (p) = C  ln ((√(y^2 +x^2 ))) + tan^(−1) ((y/x)) = C

Commented by1549442205PVT last updated on 25/Jul/20

this result don′t allow us to check  original condition easily

Answered by ajfour last updated on 24/Jul/20

let  y=tx ⇒   (dy/dx)= t+x(dt/dx)      x(dt/dx) = ((t−1)/(t+1))−((t(t+1))/(t+1)) = −(((t^2 +1))/((t+1)))  ⇒   ∫((t+1)/(t^2 +1))dt =−∫ (dx/x)           (1/2)ln ∣t^2 +1∣+tan^(−1) t+ln x = c  ⇒  (1/2)ln ∣x^2 +y^2 ∣+tan^(−1) ((y/x))=c    (no idea how to evaluate c )

Answered by 1549442205PVT last updated on 24/Jul/20

Put x=yt⇒dx=ydt+tdy⇒(dy/dx)=((y−x)/(y+x))=((1−t)/(t+1))  ((ydt+tdy)/dy)=((1+t)/(1−t))⇒((ydt)/dy)=((1−t^2 )/(1−t))⇒(dy/y)=((1−t)/(1−t^2 ))dt  (dy/y)=(dt/(1+t))⇒ln∣y∣=ln∣1+t∣+lnC(C>0)  ⇒y=C(1+t)=C(1+(x/y))  ⇔y^2 =C(x+y)⇔y^2 −Cy−Cx=0  𝚫=C^2 +4Cx⇒y=((C+(√(C^2 +4Cx)))/2)   or y=((C−(√(C^2 +4Cx)))/2)  i)If y=((C+(√(C^2 +4Cx)))/2)  then  y(0)=0⇒0=((C+C)/2)⇔C=0⇒y≡0  contradiction.  ii)If y=((C−(√(C^2 +4Cx)))/2)  then   y(0)=0⇔0=((C−C)/2) is true ∀C>0  Thus ,y=((C−(√(C^2 +4Cx)))/2)