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 ))