Solving Differential Equation

**sparky** · February 21st 2011, 02:59 PM

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

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

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

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

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

**topsquark** · February 21st 2011, 03:58 PM

Originally Posted by sparky

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

Are you asked to solve this or put it into a more standard form?

-Dan

**sparky** · February 21st 2011, 04:16 PM

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.

**sparky** · February 21st 2011, 04:17 PM

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.

**topsquark** · February 21st 2011, 04:28 PM

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

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

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

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

$2xydy - 5dy = -2xdx -y^2dx$ (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:
$(2xydy + y^2dx) - 5dy + 2xdx = 0$

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

-Dan

**sparky** · February 21st 2011, 05:47 PM

Originally Posted by topsquark

You are good for as far as you have gotten.

Ok this is very encouraging.

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

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

**topsquark** · February 21st 2011, 06:20 PM

Originally Posted by sparky

Ok this is very encouraging.

I am still trying to figure out why you grouped together $(2xydy + y^2dx)$ rather than $(2xydy-5dy)$ . 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
$(2xydy + y^2dx) = d(xy^2)$

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

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

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