Exact equation and Homogeneous function

 

Def 1

A differential expression $M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y$ is an exact differential in $R = \left\{ (x, y) \in \mathbb{R}^{2} \; | \; a<x<b, \; c<y<d \right\}$ if it corresponds to the differential of some function $f(x, y)$ defined in $R$. A $1$st order DE of the form

$M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y = 0$

is called exact equation if $M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y$ is an exact differentail in $R$

 

 

 

Thm 1 criteria of exact differential

Let $M(x, y)$ and $N(x, y)$ be continuous and have continuous first partial derivatives in $R = \left\{ (x, y) \in \mathbb{R}^{2} \; | \; a<x<b, \; c<y<d \right\}$. Then $M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y$ is an exact differential iff

$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$

 

pf)

$\Rightarrow)$

  For simplicity we assume that $M(x, y), N(x, y)$ have continuous first partial derivatives in $\mathbb{R}^{2}$.

  Then there exists some function $f(x, y)$  s.t.  $$M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y = \frac{\partial f}{\partial x} \mathrm{d}x + \frac{\partial f}{\partial y} \mathrm{d}y$$

  Therefore $$M(x, y) = \frac{\partial f}{\partial x}, \; N(x, y) = \frac{\partial f}{\partial y}$$

  and $$\frac{\partial M}{\partial y} = \frac{\partial }{\partial y}(\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial y \partial x} = \frac{\partial }{\partial x}(\frac{\partial f}{\partial y}) = \frac{\partial N}{\partial x}$$

 

$\Leftarrow)$

  Let $f(x, y) = \int M(x, y) \mathrm{d}x + g(y)$.

  At this time, $g'(y) = N(x, y) - \frac{\partial }{\partial y} \int M(x, y) \mathrm{d}x $.

 

  Claim 1 : $g(y)$ depends only on $y$ 

  It is sufficient to show $\frac{\partial }{\partial x} [g'(y)] = 0$

  $\frac{\partial }{\partial x} [g'(y)] = \frac{\partial }{\partial x} [N(x, y) - \frac{\partial }{\partial y} \int M(x, y) \mathrm{d}x ]$

                      $= \frac{\partial N}{\partial x} - \frac{\partial }{\partial y} (\frac{\partial }{\partial x} \int M(x, y) \mathrm{d}x)$

                      $= \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 0$

 

  Claim 2 : $\mathrm{d}f(x, y) = M(x, y)\mathrm{d}x + N(x, y)\mathrm{d}y$

  $\mathrm{d}f(x, y) = \frac{\partial }{\partial x}[\int M(x, y) \mathrm{d}x + g(y)] \mathrm{d}x + \frac{\partial }{\partial y}[\int M(x, y) \mathrm{d}x + g(y)] \mathrm{d}y$
                    $= M(x, y) \mathrm{d}x + [\frac{\partial }{\partial y}\int M(x, y) \mathrm{d}x + g'(y)] \mathrm{d}y$
                    $= M(x, y) \mathrm{d}x + N(x,y) \mathrm{d}y $

 

  $\therefore$  $M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y$ is an exact differential

 

 

 

Def 2

A function $f(x, y)$ is called homogeneous function of degree $\alpha$ if  $\forall t, x, y \in \mathbb{R},$

$f(tx, ty) = t^{\alpha}f(x,y)$

 

 

 

Def 3

A $1$st order DE $M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y = 0$ is said to be homogeneous if both $M(x, y)$ and $N(x, y)$ are homogeneous function of the same degree $\alpha$.

 

 

 

Solution

Idea : $u = \frac{y}{x}$  $\Rightarrow$  $\mathrm{d}y = u \mathrm{d}x + x \mathrm{d}u$

$M(x, y) = M(x, ux) = x^{\alpha} M(1, u) $

$N(x, y) = N(x, ux) = x^{\alpha} N(1, u) $

 

$M(x, y) \mathrm{d}x + N(x, y) \mathrm{d}y = 0$

     $\Rightarrow$  $x^{\alpha}M(1, u) \mathrm{d}x + x^{\alpha}N(1, u) \mathrm{d}y = 0$

     $\Rightarrow$  $x^{\alpha}M(1, u) \mathrm{d}x + x^{\alpha}N(1, u) [u \mathrm{d}x + x \mathrm{d}u ]= 0$

     $\Rightarrow$  $[M(1, u) + uN(1, u)]\mathrm{d}x + xN(1, u)\mathrm{d}u = 0$

     $\Rightarrow$  $\frac{1}{x}\mathrm{d}x = -\frac{N(1, u)}{M(1, u)+uN(1, u)}\mathrm{d}u$

It is separable.