Who can find the solution to this ODE?

[Cody Palmer]

Looking for the solution to the following ODE:
<br /> \[<br /> \frac{dy}{dx} = \frac{x^2y^2 - y}{x}<br /> \]<br />

 

[rock.freak667]

rewrite it as

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


then read about http://en.wikipedia.org/wiki/Bernoulli_differential_equation"

 

[Cody Palmer]

I had not thought of turning it into a Bernoulli equation. Actually though I found out another, more elementary way of doing it using exact differentials:
<br /> \[<br /> \frac{dy}{dx} = \frac{x^2y^2 - y}{x} \Rightarrow x dy = x^2y^2 dx - y dx<br /> \]<br />
Divide through by y^2
<br /> \[<br /> x\frac{1}{y^2} dy = x^2 dx - \frac{1}{y} dx \Rightarrow x\frac{1}{y^2} dy - (- \frac{1}{y}) dx = x^2 dx<br /> \]<br /> \[<br /> \frac{x\frac{1}{y^2} dy - (- \frac{1}{y}) dx}{x^2} = dx<br /> \]<br />
The LHS is the exact differential for the quotient \frac{-\frac{1}{y}}{x}
So we can integrate both sides to get
<br /> \[<br /> -\frac{1}{xy} = x + c \Rightarrow y=-\frac{1}{x^2 + cx}<br /> \]<br />

 

[coomast]

Nice Cody Palmer, very nice

coomast