Question Number 103759 by bemath last updated on 17/Jul/20

(x^5 +3y) dx −x dy = 0

Answered by MAB last updated on 17/Jul/20

x^5 +3y−xy′=0  xy′−3y=x^5   z=(y/x^3 )→z′=((xy′−3y)/x^4 )  x^4 z′=x^5   z′=x  z=(1/2)x^2 +c          (c∈C)  y=x^3 z=(1/2)x^5 +cx^3

Answered by john santu last updated on 17/Jul/20

x (dy/dx) −3y = x^5    (dy/dx)− (3/x)y = x^4    Integrating factor   IF u(x) = e^(∫−(3/x) dx) = e^(ln((1/x^3 )))   u(x) = (1/x^3 ) ⇒y = ((∫(1/x^3 ).x^4  dx+C)/(1/x^3 ))  y = x^3 {(1/2)x^2 +C} = (1/2)x^5 +Cx^3   (JS ⊛)

Commented byAr Brandon last updated on 17/Jul/20

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Answered by bemath last updated on 17/Jul/20