Question Number 100048 by bramlex last updated on 24/Jun/20

Answered by mr W last updated on 24/Jun/20

u=(dy/dx)  (d^2 y/dx^2 )=(du/dx)=(du/dy)×(dy/dx)=u(du/dy)  2yu(du/dy)−u^2 =1  ((2u)/(u^2 +1))du=(dy/y)  ∫((2u)/(u^2 +1))du=∫(dy/y)  ∫(1/(u^2 +1))d(u^2 +1)=∫(dy/y)  ln (u^2 +1)=ln y+C  ⇒u^2 +1=Cy  ⇒u=(dy/dx)=±(√(Cy−1))  ⇒(dy/(√(Cy−1)))=±dx  ⇒∫(dy/(√(Cy−1)))=±∫dx  (2/C)(√(Cy−1))=±x+C_1   ⇒2(√(Cy−1))=±C(x+C_1 )