0
$\begingroup$

$(xy+y^2+1)dx+(x^2+xy+1)dy=0$. 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

$\endgroup$
0
$\begingroup$

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 $$xy + \ln(x+y) = \text{constant.} $$

$\endgroup$
  • $\begingroup$ 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 ^^ $\endgroup$ – Mann May 9 '15 at 13:17
0
$\begingroup$

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 $f=e^{xy}$.

$\endgroup$
  • $\begingroup$ How did you find xy $\endgroup$ – Gizem May 9 '15 at 13:38
  • $\begingroup$ 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. $\endgroup$ – JJacquelin May 9 '15 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.