Differential Equation problem (Exact Equation)

• Oct 24th 2018, 11:05 AM
Belmond
Differential Equation problem (Exact Equation)
verify that the given differential equation is not exact. (x^2 + 2xy − y^2) dx + (y^2 + 2xy − x^2) dy = 0

Multiply the given differential equation by the integrating factor μ(x, y) = (x + y)^−2 and verify that the new equation is exact.
• Oct 24th 2018, 12:42 PM
HallsofIvy
Re: Differential Equation problem (Exact Equation)
I am surprised at this. Don't you know what "exact differential equation" and "integrating factor" mean? I ask that because you have shown no attempt of your own here.

Given a function F(x,y) of two variables, its "differential" is $\displaystyle \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial F}{\partial y}dy$. Something of the form "$\displaystyle g(x,y)dx+ h(x,y)dy$ looks like that form but may not be exact. If there exist F such that $\displaystyle \frac{\partial F}{\partial x}= f(x,y)$ and $\displaystyle \frac{\partial F}{\partial y}= g(x,y)$ then $\displaystyle \frac{\partial f}{\partial y}= \frac{\partial^2 F}{\partial xy}= \frac{\partial g}{\partial x}$. In this case, $\displaystyle f(x,y)= x^2+ 2xy- y^2$ so that $\displaystyle \frac{\partial f}{\partial y}= 2x- 2y$, And $\displaystyle g(x,y)= y^2+ 2xy- x^2$ so $\displaystyle \frac{\partial g}{\partial x}= 2y- 2x$. Those are not the same so this is NOT an exact differential.

An "integrating factor" is a function such that when you multiply both parts of the non-exact differential by that function it becomes "exact". The given differential, again, is $\displaystyle (x^2+ 2xy- y^2)dx+ (y^2+ 2xy- x^2)dy$ multiplying by $\displaystyle (x + y)^{−2}$ makes that $\displaystyle \frac{x^2+ 2xy- y^2}{(x+ y)^2}dx+ \frac{y^2+ 2xy- x^2}{(x+ y)^2}dy$. Take the derivative, as above, the derivative of $\displaystyle \frac{x^2+ 2xy- y^2}{(x+ y)^2}$ with respect to y, and the derivative of $\displaystyle \frac{y^2+ 2xy- x^2}{(x+ y)^2}$ with respect to x. If they are equal, that is now exact and $\displaystyle (x+ y)^{-2}$ is an integrating factor.