0
$\begingroup$

$ DE : (y^2)dx + (x^2-xy)dy = 0 $

$ \frac{ \delta M } {\delta y} = 2y $ and $ \frac{ \delta N } {\delta x} = 2x-y $

I've found the Integrating factor to be:

$ e^\int\ \frac{((\delta M/\delta y) - (\delta N/\delta x))}{N} \ $ = $ \frac {x-y}{x^3}\ $

However after multiplying the Integrating factor with the initial equations , they don't seem to be becoming exact.

$ \delta M * \frac {x-y}{x^3} = y^2 * \frac {x-y}{x^3} $ and $ \delta N * \frac {x-y}{x^3} = x^2-xy * \frac {x-y}{x^3} $

After differentiating the new $ \delta M $ and $ \delta N $

$ \frac{\delta M } {\delta y} = - \frac {y(3y-2x)}{x^3} = \frac {-3y^2+2xy}{x^3} $ and $ \frac{\delta N } {\delta x} = \frac {2y(x-y)}{x^3} = \frac {-2y^2 + 2xy}{x^3} $

As you can see they are not exact, even though they should be. I don't know where I've made an error exactly but it seems that it's most likely in the calculation of the Integrating factor? Any help will be greatly appreciated.

$\endgroup$
  • $\begingroup$ I do unfortunately. It's part of my differential equations coursework. $\endgroup$ – Saloni Mude Mar 27 '17 at 15:26
0
$\begingroup$

The integrating factor only works if $\dfrac{\dfrac{\partial M }{\partial y } - \dfrac{\partial N }{\partial x }}{N}$ is only in terms of $x$.

Fortunately, the equation is homogeneous. Homogeneous equations can be solved using the substitution $y = u x$ and $\dfrac{dy}{dx} = x \dfrac{du}{dx} + u$, from the Product Rule.

$y^2dx + (x^2 - xy)dy = 0$ can be rewritten as $\dfrac{dy}{dx} = \dfrac{y^2}{xy - x^2}$, which simplifies to the following separable equation:

$$x \dfrac{du}{dx} = \dfrac{1}{u-1}$$

$\endgroup$

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.