Khaled's question at Yahoo Answers regarding a Bernoulli equation

[MarkFL]

Here is the question:

Solve the given bernoulli equation ( x dy/dx + y = 1/y^2)?

Solve the given bernoulli equation ( x dy/dx + y = 1/y^2)

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello khaled,

We are given to solve:

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

Let's first multiply through by \(\displaystyle \frac{y^2}{x}\):

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

Next, let's make the substitution:

\(\displaystyle v=y^3\,\therefore\,\frac{dv}{dx}=3y^2\frac{dy}{dx}\)

and we now have:

\(\displaystyle \frac{1}{3}\frac{dv}{dx}+\frac{1}{x}v=\frac{1}{x}\)

Multiply through by 3:

\(\displaystyle \frac{dv}{dx}+\frac{3}{x}v=\frac{3}{x}\)

We now have a linear ODE in $v$. Compute the integrating factor:

\(\displaystyle \mu(x)=e^{3\int\frac{dx}{x}}=x^3\)

And the ODE becomes:

\(\displaystyle x^3\frac{dv}{dx}+3x^2v=3x^2\)

The left side may now be rewritten as:

\(\displaystyle \frac{d}{dx}\left(x^3v \right)=3x^2\)

Integrate with respect to $x$:

\(\displaystyle \int\,d\left(x^3v \right)=3\int x^2\,dx\)

\(\displaystyle x^3v=x^3+C\)

\(\displaystyle v=1+Cx^{-3}\)

Back-substitute for $v$:

\(\displaystyle y^3=1+Cx^{-3}\)

Hence, the explicit solution is:

\(\displaystyle y(x)=\sqrt[3]{1+Cx^{-3}}\)