Respuesta :

nobillionaireNobley
(x^5 + 3y)dx - xdy = 0
x^5 + 3y - xdy/dx = 0
3y - xdy/dx = -x^5
dy/dx - 3/x y = x^4

I.F. = e^∫(-3/x) = 1/x^3

(dy/dx)/x^3 - (3/x)y/x^3 = x^4/x^3
(dy/dx)/x^3 - 3/x^4 y = x
(1/x^3 y)' = x
∫(1/x^3 y)'dx = xdx
1/x^3 y = 1/2 x^2 + c
y = (x^5)/2 + cx^3

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