Question Number 103664 by byaw last updated on 16/Jul/20

Answered by bramlex last updated on 16/Jul/20

(x^2 +y^2 )dx = 2xy dy (dy/dx) = ((x^2 +y^2 )/(2xy)) ; set y = zx ⇒(dy/dx) = z + x (dz/dx) ⇔ z+x (dz/dx) = ((x^2 +z^2 x^2 )/(2x^2 z)) x (dz/dx) = ((1+z^2 )/(2z))−z =((1−z^2 )/(2z)) ((2z)/(1−z^2 )) = (dx/x) ⇒∫((d(1−z^2 ))/(1−z^2 ))=−ln∣x∣+c ln∣1−z^2 ∣ = ln∣(C/x)∣ ⇒1−z^2 = (C/x) 1−(C/x) = (y^2 /x^2 ) ⇒y^2 =x^2 −Cx ⊛

Commented bybyaw last updated on 16/Jul/20

Thank you. I am greatful.