Question Number 104669 by bramlex last updated on 23/Jul/20

(x^2 +y^2 ) dy = xy dx

Answered by bemath last updated on 23/Jul/20

set y = zx   (dy/dx) = z + x (dz/dx)  ⇒ (dy/dx) = ((xy)/(x^2 +y^2 ))  ⇒z + x (dz/dx) = ((zx^2 )/(x^2 (1+z^2 )))  z + x (dz/dx) = (z/(1+z^2 ))  x (dz/dx) = ((z−z−z^3 )/(1+z^2 ))  ((1+z^2 )/z^3 ) dz = −(dx/x)  ∫ {z^(−3) + (1/z)} dz = ln ∣(C/x)∣  −(1/(2z^2 )) = ln ∣(C/(zx))∣   (1/(2z^2 )) = ln ∣((zx)/C)∣ ⇒((zx)/C) = e^(1/(2z^2 ))   ∴ y = Ce^(1/((((2y^2 )/x^2 )))) = C^(  (x^2 /(2y^2 )))  ■

Answered by OlafThorendsen last updated on 23/Jul/20

y(x(dx/dy))−x^2  = y^2   (1/2)yx^2 −∫(1/2)x^2 dy−∫x^2 dy = (y^3 /3)+C_1   u = (1/2)∫x^2 dy  yu′−3u = (y^3 /3)+C_1   y^3 u′−3y^2 u = (y^5 /3)+C_1 y^2   ((y^3 u′−3y^2 u)/y^6 ) = (1/(3y))+(C_1 /y^4 )  (d/dy)((u/y^3 )) = (1/(3y))+(C_1 /y^4 )  (u/y^3 ) = (1/3)ln∣y∣+(C_2 /y^3 )+C_3   u = (1/3)y^3 ln∣y∣+C_3 y^3 +C_2   x^2  = 2u′ = 2y^2 ln∣y∣+((2y^2 )/3) +C_4 y^2   x^2  = 2y^2 ln∣y∣ +C_5 y^2   x(y) = ±y(√(lny^2 +C_5 ))

Answered by Dwaipayan Shikari last updated on 23/Jul/20

(x^2 +v^2 x^2 )dy=vx^2 dx  (dy/dx)=(v/(1+v^2 ))  (dv/dx)x=((v−v−v^3 )/(1+v^2 ))  ∫((1+v^2 )/v^3 )dv=∫−(dx/x)  −(1/(2v^2 ))+logv=−logx+C  (1/(2v^2 ))=log(vx)+C  (x^2 /(2y^2 ))=log(y)+log(C_1 )  y=C_2 e^(x^2 /(2y^2 ))