# Integrating factor differential equations

$(xy+y^2+1)dx+(x^2+xy+1)dy=0$. I know its not exact and I can't find the integrating factor. I use e over integrating but still not find. Do you help me I wonder what I'm doing wrong

suppose $$a(xy+y^2+1)\, dx + a(x^2+xy+1)\, dy = 0$$ is exact. then we need $a$ to satisfy $$a_y(xy+y^2+1) + a(x+2y) = a_x(x^2+xy+1) + a(2x+y).$$ rewrite this equation as $$a_x(x^2+xy+1) -a_y(xy+y^2+1) + a(x-y) = 0\tag1$$

look for a solution $a$ to $(1)$ in the form of $$a= f(x+y) \tag 2$$

subbing $(2)$ in (1), we find that $$\left(x^2 + xy+1 -xy-y^2-1\right)f'+(x-y)f = 0\to f=\frac1{x+y}$$ we can now integrate the exact differential equation $$\frac{xy+y^2+1}{x+y}\, dx + \frac{xy+x^2+1}{x+y} \,dy = 0\to \left(y+\frac1{x+y} \right)\, dy + \left(x+\frac1{x+y} \right)\, dx = 0.$$ the solution is $$xy + \ln(x+y) = \text{constant.}$$

• I actually tried expansion and looking for nice factor came up like this $ydx(x+y)+xdy(x+y)+d(x+y)=0$ I guess where it leads now ^^ – Mann May 9 '15 at 13:17

Try the integrating factor $e^{xy}$

How to find this integrating factor ?

One observe the symetry of the differential equation relatively to $x$ and $y$. This draw us to search an integrating factor on the form $f(xy)$ so that the differential equation becomes exact :

$$dF=f(xy)(xy+y^2+1)dx+f(xy)(x^2+xy+1)dy=0$$ where $\frac{dF}{dx}=f(xy)(xy+y^2+1)$ and $\frac{dF}{dx}=f(xy)(x^2+xy+1)$

The condition to be exact is : $$\frac{d}{dy}\left(f(xy)(xy+y^2+1)\right)=\frac{d}{dx}\left(f(xy)(x^2+xy+1)\right)$$ I let you continue : With $t=xy$ after simplification the relationship is reduced to $\frac{df}{dt}=f(t)$ which gives $f=e^{xy}$.

• How did you find xy – Gizem May 9 '15 at 13:38
• See the addition to my first answer. It is usal, before trying more complicated things, to try some integrating factors of the form $f(x)$, and $f(y)$ and $f(xy)$. The symetry of the equation draw to try $f(xy)$ in first place. – JJacquelin May 9 '15 at 14:13