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.

x \frac{dy}{dx} + y = \frac{1}{y^2}

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

Divide by x

\frac{dy}{dx} + \frac{y}{x} = \frac{1}{xy^2}

Multiply by y²

y^2 \frac{dy}{dx} + \frac{y^3}{x} = \frac{1}{x}

So w=y^3

\frac{dw}{dx}=3y^2\frac{dy}{dx}

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

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

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

w=1, so y^3=w=1=y

Which one can see works easily when plugging into the original formula. However the book gives a different answer of y^3=1+cx^-^3 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 \frac{d(wx^3)}{dx}=3x^2

w=1, so y^3=w=1=y

Which one can see works easily when plugging into the original formula. However the book gives a different answer of y^3=1+cx^-^3 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
\frac{d(wx^3)}{dx}=3x^2

then wx^3= ?

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...