Question Number 103483 by bemath last updated on 15/Jul/20

y^2 −u(u+y). (dy/du) = 0

Answered by Dwaipayan Shikari last updated on 15/Jul/20

y^2 =(dy/du).u(u+y)  y^2 =(dy/du)vy(vy+y)         {u=vy ,      1=(dv/du)y+(dy/du)v  1=(dy/du)v(v+1)  1=(1−(dv/du)y)(v+1)  1=v+1−(dv/du)y(v+1)  (v/(v+1))=(dv/du)((u/v))  ∫(du/u)=∫((v+1)/v^2 )dv  logu=logv−(1/v)+C  logy=−(y/u)+log(C_1 )  log((y/C_1 ))=−(y/u)  y=C_1 e^((−y)/u)

Commented byDwaipayan Shikari last updated on 15/Jul/20

I have  started my lesson on Differential equations . Kindly  check my answer

Answered by bramlex last updated on 15/Jul/20

(dy/du) = (y^2 /(u^2 +uy)) . set y = uw   (dy/du) = w + u(dw/du)  ⇔ w + u (dw/du) = ((u^2 w^2 )/(u^2 +u^2 w))  w + u (dw/du) = (w^2 /(1+w))  u (dw/du) = (w^2 /(1+w))−w = ((−w)/(1+w))  ((1+w )/w) dw = −(du/u)   ∫ ((1/w)+1)dw = −ln∣u∣+C  ln∣w∣+w = C−ln∣u∣  ln∣uw∣ +w = C  ln∣u.(y/u)∣+(y/u)=C  u ln∣y∣ +y = Cu  y = u(C−ln∣y∣) ■

Answered by OlafThorendsen last updated on 15/Jul/20

y = uY  u^2 Y^2 −u(u+uY)(u(dY/du)+Y) = 0  u^2 Y^2 −u^3 (1+Y)(dY/du)−u^2 (1+Y)Y = 0  −u^3 (1+Y)(dY/du)−u^2 Y = 0  ((1+Y)/Y)dY = −(du/u)  ln∣Y∣+Y = −ln∣u∣+C  ln∣u∣ = −ln∣Y∣−Y+C  ∣u∣ = ((Ke^(−Y) )/(∣Y∣)) = ((Ke^(−(y/u)) )/(∣(y/u)∣))  u^2  = ((Ke^(−(y/u)) )/y)