# Seperable Differential equation

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

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

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

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

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

Originally Posted by Deadstar
Ok I'll separate the variables for you...

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

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

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

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

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

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

$\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 $x$ and $y$. That's the way it already is, right?
• Jun 9th 2010, 04:31 PM
craig
Quote:

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

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

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

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

$\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 $x$ and $y$. That's the way it already is, right?

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

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

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

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

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

$\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 $x$ and $y$. That's the way it already is, right?

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

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

You are right. Thanks, Deadstar.