Thread: Solve the following ODE ..#2

1. Solve the following ODE ..#2

Hello:

Problem:
Solve the following equation :
$2y(x+y+2)dx+(y^2-x^2-4x-1)dy=0$

Solution:
I did not get it ..
not separable, not homogeneous, not exact, not linear, not bernoulli
So I tried to use the method of determinating the integrating factor ..
But this failed with me:

$\dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right)$ is not a function in x only ..

and

$\dfrac{1}{M} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right)$ is not a function in y only ..

any ideas??

2. It is an exact equation

$\dfrac{1}{M} \left( \dfrac{\partial N}{\partial y} - \dfrac{ \partial M }{ \partial x} \right)$

$\frac{(-2x-4)-(2x+4y+4)}{2xy+2y^2+4y}$ = $\frac{-4x-4y-8}{2xy+2y^2+4y}$ = $\frac{-x-y-2}{xy+y^2+2y}$ = $\frac{-(x+y)-2}{y[(x+y)+2]}$ = $\frac{-(x+y)-2}{-y[-(x+y)-2]}$

so

it equals

$\frac{-1}{y}$

Can you finish up from here?

3. Wait
In my notes & book, its only :

$\dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right)$ and it should be function in x only

&

$\dfrac{1}{M} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right)$ and it should be function in y only

There is no thing about : $\dfrac{1}{M} \left( \dfrac{\partial N}{\partial y} - \dfrac{ \partial M }{ \partial x} \right)$

any one know is the latter valid also ??
if so, can you tell me all expressions??

4. Originally Posted by Liverpool
Wait
In my notes & book, its only :

$\dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right)$ and it should be function in x only

&

$\dfrac{1}{M} \left( \dfrac{ \partial N }{ \partial x}-\dfrac{\partial M}{\partial y} \right)$ and it should be function in y only
those are the correct expressions.

5. sorry, I had just copied your latex and switched the M and N and forgot to switch the partial dy and dx when correcting the error. The rest of the work is correct though.

6. Thanks.