# Is this differential equation exact?

1. Aug 25, 2015

### whoareyou

1. The problem statement, all variables and given/known data

Identify the following differential equation as linear, separable, exact, or a combination of the three.
$$1 + \frac{1+x}{y}\frac{dy}{dx} = 0$$

2. Relevant equations

Start with $F(x,y)=C$

$\displaystyle \frac{d}{dx}(F(x,y)) = \frac{d}{dx} (C)$

$\displaystyle \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\frac{dy}{dx}= 0$

$\displaystyle M(x,y) = \frac{\partial F}{\partial x}$

$\displaystyle N(x,y) = \frac{\partial F}{\partial y}$

$\displaystyle M(x,y)dx + N(x,y)dy = 0$

3. The attempt at a solution

It is easy to show that the differential equation is both separable and linear by simply rearranging the equation:

$$\implies (1+x)\frac{dy}{dx} + y = 0 \\ \implies \frac{1}{y}dy = \frac{-1}{1+x}dx$$

The problem arises when checking whether or not the DE is exact. My prof and my lab instructor both gave conflicting answers.

One on hand, if you take the differential equation in its original form, $\displaystyle M(x,y) = 1$ and $\displaystyle N(x,y) = \frac{1+x}{y}$. Then it is obvious from testing the compatibility condition (i.e. $\displaystyle \frac{\partial M}{dy}(x,y) = \frac{\partial N}{dx}(x,y)$) that the given DE is not exact.

On the other hand, you can rearrange the DE to get $\displaystyle \frac{1}{1+x}dx+\frac{1}{y}dy = 0$. Then, if you let $\displaystyle M(x,y)=\frac{1}{1+x}$ and $\displaystyle N(x,y)=\frac{1}{y}$, it is clear that $\displaystyle \frac{\partial M}{dy} = \frac{\partial N}{dx} = 0$. Now the compatibility condition is met which implies the DE is exact (a result of the statement that a separable DE is always exact).

So which of these approaches are correct? A given DE is exact If and only if the compatibility condition holds.

A more general question of my problem would be: If you're given an equation $\displaystyle M(x,y)dx + N(x,y)dy = 0$ where the compatibility condition fails but its separable so $\displaystyle \frac{dy}{dx} = A(x)B(y) \implies A(x)dx - \frac{dy}{B(x)} = 0$ which is exact we would say that $\displaystyle A(x)dx - \frac{dy}{B(x)} = 0$ is exact. Do we also say that $\displaystyle M(x,y)dx + N(x,y)dy = 0$ is exact (even though initially, the compatibility condition did not hold at first)?

2. Aug 25, 2015

### Dick

Well, no. The equation as it stands is not exact. What you did to make it exact is called multiplying by an "integrating factor".

3. Aug 25, 2015

### whoareyou

So then the statement "all separable equations are exact" is not true?

4. Aug 25, 2015

### Dick

That's what I would say. On the other hand they have a pretty obvious integrating factor.

5. Aug 26, 2015

### Kilgour22

In general, the integrating factor you multiply the linear ODE by is $e^\int \frac{M_y - N_x}{N}dx$ or $e^\int \frac{N_x - M_y}{M}dy$ to make an inexact linear ODE exact, where $M_y = \frac{\partial M_(x,y)}{\partial y}$ and $N_x = \frac{\partial N_(x,y)}{\partial x}$. So if you REALLY wanted the given DE to be linear, separable and exact, you could! However, as it is currently it is most definitely not exact.