# Exact ODE: show that $y$ is a solution iff it is in a level set of $F$

A Differential equation of the form

$$p(x,y(x))\dot{y}(x)+q(x,y(x))=0\hspace{1cm}(1.1)$$

is called exact, if there is a differentiable function $$\mathbb{R}^2\mapsto\mathbb{R}$$ such that

$$p(x,y(x))=\dfrac{\partial F(x,y)}{\partial y}\hspace{2cm} q(x,y)=\dfrac{\partial F(x,y)}{\partial x}$$

Show that $$y$$ is a solution of $$(1.1)$$ iff it is in a level set of $$F$$.

$$"\Rightarrow"$$

Let $$y$$ be a solution of $$(1.1)\Rightarrow p(x,y)\dot{y}=-q(x,y)\hspace{1cm}(1.2)$$

$$\hspace{4,8cm}\Rightarrow\dfrac{\partial F(x,y)}{\partial y}\dot{y}=-\dfrac{\partial F(x,y)}{\partial x}$$

What I would now do is treat the partial derivatives as fraction and get that

$$\hspace{4,8cm}-\dfrac{\partial y}{\partial x}=\dot{y}$$

Which would mean that the derivative $$\dot{y}$$ is zero so $$y$$ is constant an therefore in a level set of $$F$$

I'm well aware that this is quite naive and would therefore like to know if the idea is atleast correct and how I can make it formally correct if so.

$$"\Leftarrow"$$

Let $$y$$ be an element of level set $$L_F(c)$$. $$\Rightarrow\exists c\in\mathbb{R}:F(x,y) = c$$ $$\hspace{6,8cm}\Rightarrow p(x,y)=q(x,y)=0$$

and so $$y$$ is a solution for $$(1.1)$$

Here I'm quite sure that it's correct.

• A hint: a solution $x\mapsto y(x)$ is in the level set of $F$ if and only if for each $x$ in the domain of $y$ there holds $$\frac{\mathrm{d}}{\mathrm{d}x}F(x,y(x))=0.$$ – user539887 May 16 at 6:56
• This definitely helped, because I could show that $\dfrac{\partial F(x,y)}{\partial x} = -\dfrac{\partial F(x,y)}{\partial x}$ but still only by calculating with derivatives as if they where fractions. Can you give me a hint on how to make it more formal aswell ? – Christian Singer May 16 at 7:07
• $$\frac{\mathrm{d}}{\mathrm{d}x}F(x,y(x))=\frac{\partial F(x,y(x))}{\partial x}+y'(x)\frac{\partial F(x,y(x))}{\partial y}$$ (see Total derivative). – user539887 May 16 at 7:16
• Duhh.. I should've catched that earlier! Thanks alot! – Christian Singer May 16 at 7:17

If $$y(x)$$ is a solution to the equation

$$p(x, y(x))\dot y(x) + q(x, y(x)) = 0, \tag 1$$

with

$$p(x, y(x)) = \dfrac{\partial F(x, y)}{\partial y}, \tag 2$$

$$q(x, y(x)) = \dfrac{\partial F(x, y)}{\partial x}, \tag 3$$

then the curve

$$\gamma(x) = (x, y(x)) \tag 4$$

with tangent vector

$$\dot \gamma(x) = (1, \dot y(x)) \tag 5$$

satisfies the equation

$$\dot \gamma(x) \cdot (q(x, y(x)), p(x, y(x)))$$ $$= (1, \dot y(x)) \cdot (q(x, y(x)), p(x, y(x))) = q(x, y(x)) + p(x, y(x))\dot y(x) = 0; \tag 6$$

in the light of (2) and (3) this may be written

$$\dot \gamma(x) \cdot \nabla F(x, y)$$ $$= (1, \dot y(x)) \cdot \nabla F(x, y) = (1, \dot y(x)) \cdot \left ( \dfrac{\partial F(x, y)}{\partial x}, \dfrac{\partial F(x, y)}{\partial y} \right ) = 0, \tag 7$$

and since

$$\dfrac{dF(x, y(x))}{dx} = (1, \dot y(x)) \cdot \nabla F(x, y(x)) = \dot \gamma(x) \cdot \nabla F(\gamma(x)), \tag 8$$

we see that

$$\dfrac{dF(x, y(x))}{dx} = 0, \tag 9$$

that is, the curve $$\gamma(x) = (x, y(x))$$ lies in a set of constant $$F(x, y)$$, also known as a level set of $$F$$.

Now suppose that $$y(x)$$ is a differentiable function of $$x$$ which satisfies (9); then via (8) we see that (7) binds, and thus that

$$(1, \dot y(x)) \cdot \left ( \dfrac{\partial F(x, y)}{\partial x}, \dfrac{\partial F(x, y)}{\partial y} \right ) = 0, \tag{10}$$

or

$$\dfrac{\partial F(x, y)}{\partial x} + \dfrac{\partial F(x, y)}{\partial y} \dot y(x) = 0, \tag{11}$$

and by means of (2)-(3) we immediately arrive at (1), which we see that $$y(x)$$ satisfies. Note that we have stipulated that $$y(x)$$ be differentiable in order to ensure the existence of $$\dot y(x)$$, essential if the argument is to make sense at all. In any event, that differentiable $$y(x)$$ is a solution of (1) if and only if it's graph lises in a level set of $$F(x, y)$$ has been established.

• Thanks again for this in-depth coverage! Always appreciated – Christian Singer May 19 at 10:43
• @ChristianSinger: you are more than welcome as usual, my friend. And thanks for the "acceptance". Cheers! – Robert Lewis May 19 at 16:17