Solving Bernouli Equation with x $\frac{dy}{dx}+y=\frac{1}{y^2}$

[n1person]

Recently I've been going back in my differential equation book to review some differential equation solving skills, in particular bernouli, ricatti's, and clairaut's equations; simple things enough. However when doing the exercises I have kept running into a "problem" with one question.

[tex]x \frac{dy}{dx} + y = \frac{1}{y^2}[/tex]

Now I get it into the form to use the bernouli equation:

Divide by x

[tex]\frac{dy}{dx} + \frac{y}{x} = \frac{1}{xy^2}[/tex]

Multiply by y²

[tex]y^2 \frac{dy}{dx} + \frac{y^3}{x} = \frac{1}{x}[/tex]

So [tex]w=y^3[/tex]

[tex]\frac{dw}{dx}=3y^2\frac{dy}{dx}[/tex]

So now the equation is [tex]\frac{dw}{dx} + \frac{3w}{x} = \frac{3}{x}[/tex]

So the integrating factor is [tex]e^\int^\frac{3}{x}^d^x=x^3[/tex]

So you get [tex]\frac{d(wx^3)}{dx}=3x^2[/tex]

w=1, so [tex]y^3=w=1=y[/tex]

Which one can see works easily when plugging into the original formula. However the book gives a different answer of [tex]y^3=1+cx^-^3[/tex] which I don't understand how one could get with bernouli (and my calculator gives me the same answer with deSolve)... any help?

 

[rock.freak667]

So you get [tex]\frac{d(wx^3)}{dx}=3x^2[/tex]

w=1, so [tex]y^3=w=1=y[/tex]

Which one can see works easily when plugging into the original formula. However the book gives a different answer of [tex]y^3=1+cx^-^3[/tex] which I don't understand how one could get with bernouli (and my calculator gives me the same answer with deSolve)... any help?

Now if
[tex]\frac{d(wx^3)}{dx}=3x^2[/tex]

then [itex]wx^3= ?[/itex]

When you get that, sub back w=y³ and then divide by x³

 

[n1person]

ahhhh! I forgot the constant of integration :( how silly of me...