# Math Help - Show that y(x) satisfies equation

1. ## Show that y(x) satisfies equation

Assume that the implicit function $e^{xy} = y$ determines y as a differentiable function of x in some interval. Without attempting to solve the equation for y(x), show that y(x) satisfies the differential equation $(1-xy)y' - y^2 = 0$.

I started by finding the derivative of y.
$y' = y'e^x \Rightarrow$
$e^{xy} = 1$

I am unsure of what to do next to answer the question.

EDIT: Looking at this problem more, I realized that I forgot to use the multiplication rule when finding the derivative of $xy$

2. ## Re: Show that y(x) satisfies equation

Originally Posted by Mike9182
Assume that the implicit function $e^{xy} = y$ determines y as a differentiable function of x in some interval. Without attempting to solve the equation for y(x), show that y(x) satisfies the differential equation $(1-xy)y' - y^2 = 0$.

I started by finding the derivative of y.
$y' = y'e^x \Rightarrow$
$e^{xy} = 1$

I am unsure of what to do next to answer the question.
I think your derivative is incorrect... Let $\displaystyle u = xy \implies y = e^u$, then

\displaystyle \begin{align*} \frac{du}{dx} &= x\frac{dy}{dx} + y \\ \\ \frac{dy}{du} &= e^u \\ &= e^{xy} \\ \\ \frac{dy}{dx} &= e^{xy}\left(x\frac{dy}{dx} + y\right) \\ \frac{dy}{dx} &= x\,e^{xy}\frac{dy}{dx} + y\,e^{xy} \\ \frac{dy}{dx} - x\,e^{xy}\frac{dy}{dx} &= y\,e^{xy} \\ \frac{dy}{dx}\left(1 - x\,e^{xy}\right) &= y\,e^{xy} \\ \frac{dy}{dx} &= \frac{y\,e^{xy}}{1 - x\,e^{xy}} \end{align*}

Now substitute this into the original DE.

3. ## Re: Show that y(x) satisfies equation

I substituted the result into the DE and then played around with re-arranging things, however I am still having trouble proving the equality after I substitute.

4. ## Re: Show that y(x) satisfies equation

Show us your work, and we might be able to see where the problem is.

5. ## Re: Show that y(x) satisfies equation

$(1 - xy)(y' - y^2) = 0 (Original DE) \Rightarrow$

$(1 - xe^{xy})(\frac{ye^{xy}}{1-xe^{xy}} - y^2) = 0 \Rightarrow$

$ye^{xy} - (e^{xy})^2(1-xe^{xy}) = 0 \Rightarrow$

$ye^{xy}-(e^{xy})^2 + x(e^{xy})^3 = 0 \Rightarrow$

$e^{xy}(y - e^{xy} + x(e^{xy})^2) = 0 \Rightarrow$

$e^{xy} - e^{xy} + x(e^{xy})^2 = 0 \Rightarrow$

$e^{xy}(1 - 1 + x(e^{xy}) = 0 \Rightarrow$

$e^{xy}(xe^{xy}) = 0$

6. ## Re: Show that y(x) satisfies equation

Originally Posted by Mike9182
$(1 - xy)(y' - y^2) = 0 (Original DE) \Rightarrow$
This is not the original DE. The original DE is

$(1 - xy)y' - y^{2} = 0.$

Another suggestion: instead of writing all those exponentials, I would use your implicit equation for y to get rid of them in the derivative. Less writing that way, though it should work out in either case.

$(1 - xe^{xy})(\frac{ye^{xy}}{1-xe^{xy}} - y^2) = 0 \Rightarrow$

$ye^{xy} - (e^{xy})^2(1-xe^{xy}) = 0 \Rightarrow$

$ye^{xy}-(e^{xy})^2 + x(e^{xy})^3 = 0 \Rightarrow$

$e^{xy}(y - e^{xy} + x(e^{xy})^2) = 0 \Rightarrow$

$e^{xy} - e^{xy} + x(e^{xy})^2 = 0 \Rightarrow$

$e^{xy}(1 - 1 + x(e^{xy}) = 0 \Rightarrow$

$e^{xy}(xe^{xy}) = 0$