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Find the general solution of The ff. D.E

bergausstein

Active member
Jul 30, 2013
191
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

Administrator
Staff member
Feb 24, 2012
13,775
Look at your second option again...you should find it is a function of $y$ alone. :D
 

bergausstein

Active member
Jul 30, 2013
191
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

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

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

I'm confused! help!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Jul 30, 2013
191
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$
Correct! Now look again at the expression you want to use to compute your integrating factor...what do you find?
 

bergausstein

Active member
Jul 30, 2013
191

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

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Try the second option...
 

bergausstein

Active member
Jul 30, 2013
191
both options are not a function of just x and y.

as I stated in my OP. :(
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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