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)