# Seperable Differential equation

• Jun 9th 2010, 03:04 PM
Hibachi
Seperable Differential equation
Hey, I got stuck on the following differential equation. Given $\displaystyle y = 0$ when $\displaystyle x = 2$, solve $\displaystyle \dfrac{x}{y}\cdot\dfrac{dx}{dy} = \dfrac{x^2+1}{y^2+1}$.

I can't separate the variables and I couldn't turn it into a form that I can take an integrating factor.
• Jun 9th 2010, 03:33 PM
Ok I'll separate the variables for you...

$\displaystyle \dfrac{x}{y}\cdot\dfrac{dx}{dy} = \dfrac{x^2+1}{y^2+1}$

$\displaystyle => \dfrac{x}{x^2 + 1}dx = \dfrac{y}{y^2+1} dy$

• Jun 9th 2010, 03:55 PM
Hibachi
Quote:

Ok I'll separate the variables for you...

$\displaystyle \dfrac{x}{y}\cdot\dfrac{dx}{dy} = \dfrac{x^2+1}{y^2+1}$

$\displaystyle => \dfrac{x}{x^2 + 1}dx = \dfrac{y}{y^2+1} dy$

So $\displaystyle \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C$

when $\displaystyle y = 0$, $\displaystyle x = 2$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(3) = C$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.$

The question said in the begening express teh solution as an implicit equation relating $\displaystyle x$ and $\displaystyle y$. That's the way it already is, right?
• Jun 9th 2010, 04:31 PM
craig
Quote:

Originally Posted by Hibachi
So $\displaystyle \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C$

when $\displaystyle y = 0$, $\displaystyle x = 2$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(3) = C$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.$

The question said in the begening express teh solution as an implicit equation relating $\displaystyle x$ and $\displaystyle y$. That's the way it already is, right?

Yes :)
• Jun 10th 2010, 03:09 AM
Quote:

Originally Posted by Hibachi
So $\displaystyle \int\dfrac{x}{x^2 + 1}dx = \int\dfrac{y}{y^2+1} dy$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+C$

when $\displaystyle y = 0$, $\displaystyle x = 2$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(3) = C$

$\displaystyle \therefore \;\;\; \dfrac{1}{2}\log(x^2+1) = \dfrac{1}{2}\log(y^2+1)+\log\sqrt{3}.$

The question said in the begening express teh solution as an implicit equation relating $\displaystyle x$ and $\displaystyle y$. That's the way it already is, right?

It's the right method yes but I think it should be $\displaystyle \log(5)$ should it not? Since the $\displaystyle x$ is squared.
• Jun 10th 2010, 09:42 AM
Hibachi
Quote:

It's the right method yes but I think it should be $\displaystyle \log(5)$ should it not? Since the $\displaystyle x$ is squared.