# How to prove that $y'=e^{xy}$ is not separable?

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.

• 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. Oct 31, 2019 at 18:22

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.

$$x = f(y)$$

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

$$1 = 0$$

• 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?
– jte
Oct 31, 2019 at 18:12
• I've added more to my answer. Let me know if that helps. Oct 31, 2019 at 18:49
• Yes! Thank you for the clarification.
– jte
Nov 1, 2019 at 19:10

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.

• 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?
– jte
Oct 31, 2019 at 18:39