# Thread: Help with diff eqn

1. ## Help with diff eqn

How can I proceed with this? solve the diff eqn: $\left (2y^2 - 4x + 5 \right )dx = \left (4 - 2y + 4xy \right )dy$

$\frac{dy}{dx} = \frac{2y^2 -4x + 5}{4 -2y + 4xy}$

Doesn't seem I can use quotient rule here.

..............?

2. Forgive me if I err along the line, but PH or someone will be along to correct me if I wonder.

It's been awhile.

Now, check to see if your DE is exact.

$(2y^{2}-4x+5)dx=M(x,y)$

$(4-2y+4xy)dy=N(x,y)$

$M_{y}=4y=N_{x}$

There exists a function f(x,y), such that:

$\frac{\partial{f}}{\partial{x}}=2y^{2}-4x+5$

$\frac{\partial{f}}{\partial{y}}=4-2y+4xy$

From the 1st equation, after integrating:

$f(x,y)=2xy^{2}+5x-2x^{2}+g(y)$

Take the partial derivative with respect to y of the last expression and set the result equal to N(x,y):

$\frac{\partial{f}}{\partial{y}}=4xy+g'(y)=4-2y+4xy$

$g'(y)=4-2y$

and $g(y)=4y-y^{2}$

$f(x,y)=2xy^{2}+5x-2x^{2}+4y-y^{2}$

Therefore, $\frac{\partial{f}}{\partial{x}}=2y^{2}-4x+5$

$\frac{\partial{f}}{\partial{y}}=4-2y+4xy$

So, $f(x,y)=2xy^{2}+5x-2x^{2}+4y-y^{2}$

Check to make sure I didn't boo-boo.

3. I do not think so.

First it needs to have the form,
$M(x,y)+N(x,y)y'=0$
And in this case,
$\frac{\partial M}{\partial y}=4y$
$\frac{\partial N}{\partial x}=-4y$
Thus, this is not an exact differencial equation.
Also, I do not think it is possible to find an integrating factor.

But what I think happened is that the poster wrote a wrong question. And really meant what you wrote.

4. I kinda thought, but figured I'd take a stab, in case.

5. Thanks for your help. This is how I got the question, Solve diff eqn -

$\left (2y^2 - 4x + 5 \right )dx = \left (4 - 2y + 4xy \right )dy$

6. Originally Posted by ashura
Thanks for your help. This is how I got the question, Solve diff eqn -

$\left (2y^2 - 4x + 5 \right )dx = \left (4 - 2y + 4xy \right )dy$
As I said, this is not an exact differencial equation. You cannot even find an integrating factor that will transform this into an exact differencial equation. This is not a homogenous equation because the polynomials are not homogenous of the same degree.

Maybe, there is something else that you need to do, but it certainly is not one of the standard techiniques.