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Very new to this subject. I know that the definition for separable functions is that if it can be written as follows $$ y'=g(x)h(y) $$ then it is separable. There are many functions such as $y'=x+y$ or $y'=e^{xy}$ where you can see that it is not separable but how to actually prove that?

I tried to use the definition and write $$ e^{xy}=g(x)h(y) $$ but could not find a clear contradiction.

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    $\begingroup$ Take $x=0$ to see $h$ must be constant. Take $y=0$ to see $g$ must be constant. But then $e^{xy}$ must be constant. $\endgroup$ Oct 31, 2019 at 18:22

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We will suppose that,

$$e^{xy} = g(x)h(y),$$

in hopes of reaching a contradiction. Now take the partial derivative with respect to $y$.

$$ x e^{xy} = g(x) h'(y)$$

$$ x g(x)h(y) = g(x) h'(y)$$

Note that $g(x)$ cannot be zero for any value of $x$ because $e^{xy}$ is nonzero for all $x$ and $y$. This means we can divide both sides by $g(x)$ cancelling it from the equation. The same reasoning demonstrates that $h(y)$ is not zero for any $y$ and we can therefore also divide by $h(y)$.

$$ x h(y) = h'(y)$$

$$ x = \frac{h'(y)}{h(y)}$$

Now on the right hand side we have a function of only $y$ and on the left hand side we have a function of only $x$ which is contradictory unless both sides equal a constant value; but the left hand side is not a constant value so we have arrived at a contradiction.

Therefore the initial assumption that $e^{xy}$ can be written as $g(x)h(y)$ is not true.


Why is it a contradiction if we have a fucntion on only $y$ on the right and $x$ on the left?

$$ f(x) = g(y) $$

Well if both functions are equal to the same constant there is no contradiction. However in our case the left hand side wasn't a constant, in particular we had $f(x) = x$.

$$ x = g(y) $$

Now lets plug in two ordered pairs for $(x,y)$. Lets choose $(1,2)$ and $(2,2)$. This would produce the following equations.

$$ 1 = g(2) \qquad 2 = g(2)$$

which gives us two values for $g(2)$. However since $g$ is a function there is only one possible value for $g(2)$. This would require $1=2$ which is a contradiction because $1\neq 2$.


Another way to see it is to take the partial derivative with respect to $x$ on both sides of the equation.

$$ x = f(y)$$

$$ \frac{\partial x}{\partial x} = \frac{\partial f(y)}{\partial x}$$

$$1 = 0$$

This is a contradiction.

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  • $\begingroup$ All I'm trying to figure out now is that why is it a contradiction to have a function of only $x$ on the left and a function of only $y$ on the right? $\endgroup$
    – jte
    Oct 31, 2019 at 18:12
  • $\begingroup$ I've added more to my answer. Let me know if that helps. $\endgroup$
    – Spencer
    Oct 31, 2019 at 18:49
  • $\begingroup$ Yes! Thank you for the clarification. $\endgroup$
    – jte
    Nov 1, 2019 at 19:10
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In this case it is clear that $g(x)\ne0$ for all $x$. Consider two different values of $x$, say $a$ and $b$. Then, if $g$ and $h$ exist, we get:

$$h(y) = C_a e^{ay} = C_b e^{by}\,,$$

with $C_a=1/g(a)$ and $C_b=1/g(b)$. Taking $\ln$ in the previous equation we get:

$$ ay+\ln(C_a) = by+\ln(C_b)\,,$$

which is impossible if $a\ne b$ as required.

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  • $\begingroup$ So this should be true for all $y$ and solving for $y$ gives $y=\frac{ln(C_b)-ln(C_a)}{a-b}$ which is constant making it impossible? $\endgroup$
    – jte
    Oct 31, 2019 at 18:39

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