Find the general solution of The ff. D.E

[bergausstein]

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.

can you please help me how to go about solving this problem thanks!

 

[MarkFL]

Look at your second option again...you should find it is a function of $y$ alone. :D

 

[bergausstein]

Look at your second option again...you should find it is a function of $y$ alone. :D

what problem are you talking about? 1 or 2?

 

[MarkFL]

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. :D

 

[bergausstein]

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

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

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

I'm confused! help!

 

[MarkFL]

You aren't differentiating correctly:

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

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

 

[bergausstein]

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

 

[MarkFL]

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

Correct! Now look again at the expression you want to use to compute your integrating factor...what do you find?

 

[bergausstein]

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

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

 

[MarkFL]

Try the second option...

 

[bergausstein]

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

as I stated in my OP. :(

 

[MarkFL]

Let's take a look...

\(\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}=\frac{(6x-4y+3)-(2x-2y+1)}{2xy-y^2+y}=\frac{4x-2y+2}{y(2x-y+1)}\)

Now factor the numerator...:D