Question Number 124763 by Mammadli last updated on 05/Dec/20

Commented by Mammadli last updated on 05/Dec/20
Sorry dear ser..
1. ydx-(x³y+x)dy=0
Answered by benjo_mathlover last updated on 05/Dec/20

Commented by Mammadli last updated on 05/Dec/20
thk you dear ser, please:
ydx-(x³y+x)dy=0
Answered by liberty last updated on 06/Dec/20
![(1) y dx −(x^3 y+x)dy=0 let x = yv ⇒dx = v dy + y dv ⇔y(v dy +y dv)−(y^3 v^3 +yv)dy=0 ⇔ y^2 dv−y^3 v^3 dy = 0 ⇔ dv = y v^3 dy ; (dv/v^3 ) = y dy ∫ (dv/v^3 ) = ∫ y dy ; −(1/(2v^2 )) = (1/2)y^2 +c (y^2 /x^2 ) = −y^2 +C ; [ −2c = C ] y^2 ((1/x^2 )+1) = C ; y^2 = ((Cx^2 )/(x^2 +1)) y = ± ((λx)/( (√(x^2 +1)))) ; [ λ = (√C) ]](https://www.tinkutara.com/question/Q124783.png)