1. ## Solving Differential Equation

Hi,

I am struggling to finish this question and would love some guidance please:

Question: Solve for the following differential equation for the original equation:$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

$\displaystyle \frac{dy}{dx}= \frac{-2x-y^2}{2xy-5}$

$\displaystyle (2xy-5)dy = (-2x-y^2)dx$

$\displaystyle 2xydy - 5dy = -2xdx -y^2dx$ (assuming that I'm correct so far, what do I do now?)

2. Originally Posted by sparky
Question: Solve for the following differential equation for the original equation:$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$
Are you asked to solve this or put it into a more standard form?

-Dan

3. Originally Posted by topsquark
Are you asked to solve this or put it into a more standard form?

-Dan
Thanks for the reply. I'm asked to solve it "for the original equation". This sounds a little confusing to me.

4. Originally Posted by topsquark
Are you asked to solve this or put it into a more standard form?

-Dan
Thanks for the reply. I'm asked to solve it "for the original equation". This sounds a little confusing to me.

5. Originally Posted by sparky
Hi,

I am struggling to finish this question and would love some guidance please:

Question: Solve for the following differential equation for the original equation:$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

$\displaystyle \frac{dy}{dx}= \frac{-2x-y^2}{2xy-5}$

$\displaystyle (2xy-5)dy = (-2x-y^2)dx$

$\displaystyle 2xydy - 5dy = -2xdx -y^2dx$ (assuming that I'm correct so far, what do I do now?)
Okay, I saw a similar example in one of your other posts. You are good for as far as you have gotten. Regroup:
$\displaystyle (2xydy + y^2dx) - 5dy + 2xdx = 0$

The $\displaystyle 2xy dy + y^2 dx$ is a perfect differential. What is it?

-Dan

6. Originally Posted by topsquark
You are good for as far as you have gotten.
Ok this is very encouraging.

The $\displaystyle 2xy dy + y^2 dx$ is a perfect differential. What is it?
I am still trying to figure out why you grouped together $\displaystyle (2xydy + y^2dx)$ rather than $\displaystyle (2xydy-5dy)$. Is it because we of differentiating with respect to y (or is it x, I'm a little confused I must admit)?

7. Originally Posted by sparky
Ok this is very encouraging.

I am still trying to figure out why you grouped together $\displaystyle (2xydy + y^2dx)$ rather than $\displaystyle (2xydy-5dy)$. Is it because we of differentiating with respect to y (or is it x, I'm a little confused I must admit)?
Acutally I'm going to integrate over something different. Note that
$\displaystyle (2xydy + y^2dx) = d(xy^2)$

So when it is time to take integrals I will do it like this:
$\displaystyle \displaystyle \int d(xy^2) = xy^2 + C$

The reason why I grouped it the way I did was so I could get rid of the y dx and x dy terms.

-Dan