Integrating factos - first order DE

**Amitromem** · January 11th 2013, 06:24 AM

Hi All,
Could not solve this one: 12-x-3y+(x-y+4)y'=0 , y(0)=1
I tried to find an integrating factor in order to make it an exact equation but did not succeed.
would appreciate your help!

**Cbarker1** · January 11th 2013, 07:02 AM

Put the differential equation into standard form. then Begin to find the integrating factor.

**Amitromem** · January 11th 2013, 01:23 PM

If I understand correctly the standard form has to be: y' + P(x)*y=Q(x) and specifically: y'-3y/(x-y+4)=(x-12)/(x-y+4)
but since there is a polynomial in the denominator of P(x) which includes y this is not the standard form.
Right?

**Prove It** · January 12th 2013, 05:21 PM

I believe the standard form Cbarker was referring to is actually $\displaystyle \begin{align*} f(x,y)\,dx + g(x,y)\,dy = C \end{align*}$ , which you can then find an integrating factor for to make the equation an exact equation.

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