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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.

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  • 2
    $\begingroup$ 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 $$\frac{\mathrm{d}}{\mathrm{d}x}F(x,y(x))=0.$$ $\endgroup$ – user539887 May 16 at 6:56
  • $\begingroup$ 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 ? $\endgroup$ – Christian Singer May 16 at 7:07
  • $\begingroup$ $$\frac{\mathrm{d}}{\mathrm{d}x}F(x,y(x))=\frac{\partial F(x,y(x))}{\partial x}+y'(x)\frac{\partial F(x,y(x))}{\partial y}$$ (see Total derivative). $\endgroup$ – user539887 May 16 at 7:16
  • $\begingroup$ Duhh.. I should've catched that earlier! Thanks alot! $\endgroup$ – Christian Singer May 16 at 7:17
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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.

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

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