Integrating factor on Exact, Homogeneous DE

**cheme** · September 12th 2011, 03:26 PM

Problem: $y*(1+x)+2xy'=0$

I have rearranged the equation to put it in exact form, resulting in the equation:
$(y+xy)dx+2xdy=0$

After that I found my $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$

This leads to a complicated integrating factor of $e^{-\frac{1}{2}(ln|x|-x)}$

Am I on the correct track?

**alexmahone** · September 12th 2011, 03:40 PM

Originally Posted by cheme

Problem: $y*(1+x)+2xy'=0$

I have rearranged the equation to put it in exact form, resulting in the equation:
$(y+xy)dx+2xdy=0$

After that I found my $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$

This leads to a complicated integrating factor of $e^{-\frac{1}{2}(ln|x|-x)}$

Am I on the correct track?

Dividing throughout by $2x$ , we get

$y'+\left(\frac{1+x}{2x}\right)y=0$

The integrating factor is:

$exp\left[\int \left(\frac{1+x}{2x}\right)dx\right]$

$=exp\left[\int \left(\frac{1}{2x}+\frac{1}{2}\right) dx\right]$

$=exp\left(\frac{1}{2}ln\ x+\frac{x}{2}\right)$

$=exp\left(\frac{1}{2}ln\ x\right)*exp\left(\frac{x}{2}\right)$

$=\sqrt{x}e^{x/2}$

(Note that this isn't an exact differential equation as $\frac{\partial M}{\partial y} \neq\frac{\partial N}{\partial x}$ .)

**cheme** · September 12th 2011, 07:21 PM

I did not see to divide through by 2x. If I do that, isn't it just a separable equation?

**alexmahone** · September 12th 2011, 07:23 PM

Originally Posted by cheme

I did not see to divide through by 2x. If I do that, isn't it just a separable equation?

Yes, this is a separable equation.

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