# Integrating factos - first order DE

• Jan 11th 2013, 06:24 AM
Amitromem
Integrating factos - first order DE
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.
• Jan 11th 2013, 07:02 AM
Cbarker1
Re: Integrating factos - first order DE
Put the differential equation into standard form. then Begin to find the integrating factor.
• Jan 11th 2013, 01:23 PM
Amitromem
Re: Integrating factos - first order DE
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?
• Jan 12th 2013, 05:21 PM
Prove It
Re: Integrating factos - first order DE
I believe the standard form Cbarker was referring to is actually \displaystyle \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.