# Find the general solution of The ff. D.E

#### bergausstein

##### Active member
Find the general solution of The ff. D.E

1.$\displaystyle (2xy-y^2+y)dx+(3x^2-4xy+3x)dy=0$

2. $\displaystyle (x^2+y^2+1)dx+x(x-2y)dy=0$

i tried both of them using

$\displaystyle \frac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}$

and

$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

but none of them is a function of just x or just y.

#### MarkFL

Staff member
Look at your second option again...you should find it is a function of $y$ alone. #### bergausstein

##### Active member
Look at your second option again...you should find it is a function of $y$ alone. what problem are you talking about? 1 or 2?

#### MarkFL

Staff member
what problem are you talking about? 1 or 2?
I'm sorry, I am referring to the first problem. I wanted to get that one squared away before looking at the second one. #### bergausstein

##### Active member
$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

$\dfrac{\partial N}{\partial x}=-4xy$ and $\dfrac{\partial M}{\partial y}=2y$

$\frac{-4xy-2y}{2y}=y-4x$

I'm confused! help!

#### MarkFL

Staff member
You aren't differentiating correctly:

$$\displaystyle \frac{\partial M}{\partial y}=2x-2y+1$$

Can you find $$\displaystyle \frac{\partial N}{\partial x}$$ ?

#### bergausstein

##### Active member
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$

#### MarkFL

Staff member
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$
Correct! Now look again at the expression you want to use to compute your integrating factor...what do you find?

#### bergausstein

##### Active member

$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

$\frac{6x-4x+3-(2x-2y+1)}{2xy-y^2+y}=\frac{2y+2}{2xy-y^2+1}$ there's still x here!

Last edited:

#### MarkFL

Staff member
Try the second option...

#### bergausstein

##### Active member
both options are not a function of just x and y.

as I stated in my OP. #### MarkFL

$$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}=\frac{(6x-4y+3)-(2x-2y+1)}{2xy-y^2+y}=\frac{4x-2y+2}{y(2x-y+1)}$$
Now factor the numerator... 