bernoulli equation?

Question

dy/dx-3y/x=x^4y^(1/3)

how to slove?
what type of this equation
if i divide by x^4 it will bcm bernoulli equation
bt there is dy/dx nt dx/dy.
so how? pls help me.

Answer 1

The key is not to divide by x^4. First divide by y^(1/3),
y^(-1/3)dy/dx-3y^(2/3)/x=x^4.
Let z=y^(2/3), (2/3)y^(-1/3)dy=dy^(2/3)=dz.
Substituting in, 3dz/2dx-3z/x=x^4, dz/dx-2z/x=3x^4/2.
Integrating factor is e^{-2 Int(1/x)dx }=x^(-2).
Multiply by it, dz/x^2dx-2z/x^3=3x^2.
(d/dx)[z/x^2]=3x^2.
Integrating with dx, z/x^2=Int(3x^2)dx=x^3.
z=x^5. Substitute back, y^(2/3)=x^5. Finally, y=x^(15/2).

Answer 2

Divide by y^(1/3), now solve.